id stringlengths 16 16 | problem_id stringlengths 16 16 | style stringclasses 5
values | instruction stringclasses 30
values | input stringlengths 47 2.18k | output stringlengths 3 1.67k | final_answer stringlengths 1 580 | metadata unknown |
|---|---|---|---|---|---|---|---|
2ed4075b6fec43ca | 0192067c9aa035bb | tutor | Write a complete step-by-step solution and conclude with the final answer. | An urn has 25 red balls and 18 blue balls. If 12 balls are drawn without replacement, what is the probability that exactly 5 are red? | 1) Use a hypergeometric counting argument.
2) Total equally likely samples: C(43,12).
3) Favorable samples: choose k reds and n−k blues: C(25,5)·C(18,7).
4) Compute totals: C(43,12) = 15338678264, and favorable = 1690809120.
5) Divide favorable by total and simplify.
Final answer: $\frac{136620}{1239389}$ | \frac{136620}{1239389} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
9115a7d89f023d39 | 0192067c9aa035bb | verifier | Compute the answer and include a short verification step. Give the final answer. | An urn has 25 red balls and 18 blue balls. If 12 balls are drawn without replacement, what is the probability that exactly 5 are red? | Use a hypergeometric counting argument.
Total equally likely samples: C(43,12).
Favorable samples: choose k reds and n−k blues: C(25,5)·C(18,7).
Compute totals: C(43,12) = 15338678264, and favorable = 1690809120.
Divide favorable by total and simplify.
Check: computed probability simplifies to $\frac{136620}{1239389}$.... | \frac{136620}{1239389} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
ea219218d9d80bb9 | ae986cdff3fe460c | tutor | Give a detailed derivation, showing intermediate expressions, and then the final answer. | An urn has 10 red balls and 10 blue balls. If 6 balls are drawn without replacement, what is the probability that exactly 2 are red? | 1) Use a hypergeometric counting argument.
2) Total equally likely samples: C(20,6).
3) Favorable samples: choose k reds and n−k blues: C(10,2)·C(10,4).
4) Compute totals: C(20,6) = 38760, and favorable = 9450.
5) Divide favorable by total and simplify.
Final answer: $\frac{315}{1292}$ | \frac{315}{1292} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
5ed422ee64b333e8 | ae986cdff3fe460c | concise | Solve with minimal but sufficient steps, then provide the final answer. | An urn has 10 red balls and 10 blue balls. If 6 balls are drawn without replacement, what is the probability that exactly 2 are red? | Use a hypergeometric counting argument.
Total equally likely samples: C(20,6).
Favorable samples: choose k reds and n−k blues: C(10,2)·C(10,4).
Final answer: $\frac{315}{1292}$ | \frac{315}{1292} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
01cae1005f0cbcd5 | f29af530ecffa939 | tutor | Solve like a tutor: derive step-by-step and finish with the final answer. | A coin lands heads with probability p=$\frac{3}{5}$. If it is flipped 6 times, what is the probability of exactly 3 heads? | 1) Let X be the number of heads in n independent flips. Then X ~ Bin(n,p).
2) Use $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
3) Substitute n=6, k=3, p=$\frac{3}{5}$ and simplify.
4) Here C(6,3) = 20.
Final answer: $\frac{864}{3125}$ | \frac{864}{3125} | {
"format": "symbolic",
"domain": "probability",
"topic": "binomial_probability",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
806549846ad44fc3 | f29af530ecffa939 | concise | Give a short solution with the key steps and the final answer. | A coin lands heads with probability p=$\frac{3}{5}$. If it is flipped 6 times, what is the probability of exactly 3 heads? | Let X be the number of heads in n independent flips. Then X ~ Bin(n,p).
Use $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
Substitute n=6, k=3, p=$\frac{3}{5}$ and simplify.
Final answer: $\frac{864}{3125}$ | \frac{864}{3125} | {
"format": "symbolic",
"domain": "probability",
"topic": "binomial_probability",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
efbca7816940b6fc | 391bf82ada6159a6 | concise | Solve and give a compact solution, then the final answer. | Solve the congruences: x≡8 (mod 11), x≡1 (mod 7). Give the solution modulo 77. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 77.
Compute the CRT solution to get x ≡ 8 (mod 77).
Final answer: $x\equiv 8\pmod{77}$ | x\equiv 8\pmod{77} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
9d0771af38984635 | 391bf82ada6159a6 | tutor | Solve step-by-step like a math tutor. Show intermediate calculations and end with a final answer. | Solve the congruences: x≡8 (mod 11), x≡1 (mod 7). Give the solution modulo 77. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 77.
3) Compute the CRT solution to get x ≡ 8 (mod 77).
Final answer: $x\equiv 8\pmod{77}$ | x\equiv 8\pmod{77} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
928516f591f8b77e | 63d42fd31aa3bc5b | tutor | Solve step-by-step like a math tutor. Show intermediate calculations and end with a final answer. | A coin lands heads with probability p=$\frac{3}{8}$. If it is flipped 21 times, what is the probability of exactly 7 heads? | 1) Let X be the number of heads in n independent flips. Then X ~ Bin(n,p).
2) Use $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
3) Substitute n=21, k=7, p=$\frac{3}{8}$ and simplify.
4) Here C(21,7) = 116280.
Final answer: $\frac{194018829345703125}{1152921504606846976}$ | \frac{194018829345703125}{1152921504606846976} | {
"format": "symbolic",
"domain": "probability",
"topic": "binomial_probability",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
8f04a568558cf00b | 63d42fd31aa3bc5b | verifier | Show the solution, then validate it with a quick check. End with the final answer. | A coin lands heads with probability p=$\frac{3}{8}$. If it is flipped 21 times, what is the probability of exactly 7 heads? | Let X be the number of heads in n independent flips. Then X ~ Bin(n,p).
Use $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
Substitute n=21, k=7, p=$\frac{3}{8}$ and simplify.
Here C(21,7) = 116280.
Check: expression simplifies to $\frac{194018829345703125}{1152921504606846976}$.
Final answer: $\frac{194018829345703125}{115292150... | \frac{194018829345703125}{1152921504606846976} | {
"format": "symbolic",
"domain": "probability",
"topic": "binomial_probability",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
fd213cabdbabbd7f | 5dc4cd35c84c7b79 | tutor | Solve carefully and do not skip algebra/arithmetic steps. End with the final answer. | An urn has 30 red balls and 15 blue balls. If 8 balls are drawn without replacement, what is the probability that exactly 2 are red? | 1) Use a hypergeometric counting argument.
2) Total equally likely samples: C(45,8).
3) Favorable samples: choose k reds and n−k blues: C(30,2)·C(15,6).
4) Compute totals: C(45,8) = 215553195, and favorable = 2177175.
5) Divide favorable by total and simplify.
Final answer: $\frac{1015}{100491}$ | \frac{1015}{100491} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
ab6757378dbb566a | 5dc4cd35c84c7b79 | concise | Provide a brief derivation and the final answer. | An urn has 30 red balls and 15 blue balls. If 8 balls are drawn without replacement, what is the probability that exactly 2 are red? | Use a hypergeometric counting argument.
Total equally likely samples: C(45,8).
Favorable samples: choose k reds and n−k blues: C(30,2)·C(15,6).
Final answer: $\frac{1015}{100491}$ | \frac{1015}{100491} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
648238482905ecb7 | 7e785b8444c11f2a | tutor | Solve step-by-step like a math tutor. Show intermediate calculations and end with a final answer. | Solve the congruences: x≡3 (mod 4), x≡4 (mod 11). Give the solution modulo 44. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 44.
3) Compute the CRT solution to get x ≡ 15 (mod 44).
Final answer: $x\equiv 15\pmod{44}$ | x\equiv 15\pmod{44} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
ac5b1a2090ad9854 | 7e785b8444c11f2a | concise | Solve with minimal but sufficient steps, then provide the final answer. | Solve the congruences: x≡3 (mod 4), x≡4 (mod 11). Give the solution modulo 44. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 44.
Compute the CRT solution to get x ≡ 15 (mod 44).
Final answer: $x\equiv 15\pmod{44}$ | x\equiv 15\pmod{44} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
913cf6c204438f14 | 297acbbd73f89cdd | tutor | Provide a detailed solution with numbered steps and a final answer line. | Solve the congruences: x≡2 (mod 5), x≡3 (mod 11). Give the solution modulo 55. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 55.
3) Compute the CRT solution to get x ≡ 47 (mod 55).
Final answer: $x\equiv 47\pmod{55}$ | x\equiv 47\pmod{55} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
d746d43d4e6a7ff3 | 297acbbd73f89cdd | concise | Compute the result with a short explanation and give the final answer. | Solve the congruences: x≡2 (mod 5), x≡3 (mod 11). Give the solution modulo 55. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 55.
Compute the CRT solution to get x ≡ 47 (mod 55).
Final answer: $x\equiv 47\pmod{55}$ | x\equiv 47\pmod{55} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
f02886d897917ba7 | a1ee05b52a7bc7b1 | concise | Solve and give a compact solution, then the final answer. | Solve the congruences: x≡4 (mod 11), x≡6 (mod 10). Give the solution modulo 110. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 110.
Compute the CRT solution to get x ≡ 26 (mod 110).
Final answer: $x\equiv 26\pmod{110}$ | x\equiv 26\pmod{110} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
a74f0404f32699a2 | a1ee05b52a7bc7b1 | tutor | Show your work in clear steps (with intermediate values) and finish with a final answer. | Solve the congruences: x≡4 (mod 11), x≡6 (mod 10). Give the solution modulo 110. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 110.
3) Compute the CRT solution to get x ≡ 26 (mod 110).
Final answer: $x\equiv 26\pmod{110}$ | x\equiv 26\pmod{110} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
08bffa1716633bb6 | 62d8adacb7505d93 | tutor | Provide a thorough solution as if teaching, including intermediate computations, then give the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_5=37$? | 1) We are counting nonnegative integer solutions to a sum constraint.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=37 and k=5.
4) Compute C(41,4) = 101270.
Final answer: $101270$ | 101270 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
9ce5153d324c62c5 | 660f83bc358400a6 | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Differentiate with respect to x:
$f(x)=4 x^{3} - 3 x^{2} - 2$.
Student solution:
1) Differentiate term-by-term.
2) Claim: one term $\frac{d}{dx}(4 x^{3}) = 12 x^{3}$.
3) So $f'(x) = 12 x^{3} - 6 x$. | First incorrect step: 2
The error is differentiating a power: the exponent must decrease by 1.
Correct: $\frac{d}{dx}(4 x^{3}) = 12 x^{2}$.
Therefore the correct derivative is $f'(x) = 6 x \left(2 x - 1\right)$.
Final answer: $6 x \left(2 x - 1\right)$ | 6 x \left(2 x - 1\right) | {
"format": "symbolic",
"domain": "calculus",
"topic": "derivative",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
316ac139ddd765e6 | 5b38505b62b60b20 | tutor | Show a full solution with explicit computations and a final answer line. | An urn has 11 red balls and 10 blue balls. If 6 balls are drawn without replacement, what is the probability that exactly 2 are red? | 1) Use a hypergeometric counting argument.
2) Total equally likely samples: C(21,6).
3) Favorable samples: choose k reds and n−k blues: C(11,2)·C(10,4).
4) Compute totals: C(21,6) = 54264, and favorable = 11550.
5) Divide favorable by total and simplify.
Final answer: $\frac{275}{1292}$ | \frac{275}{1292} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
51ef633db56f7fc0 | 5b38505b62b60b20 | concise | Provide a brief derivation and the final answer. | An urn has 11 red balls and 10 blue balls. If 6 balls are drawn without replacement, what is the probability that exactly 2 are red? | Use a hypergeometric counting argument.
Total equally likely samples: C(21,6).
Favorable samples: choose k reds and n−k blues: C(11,2)·C(10,4).
Final answer: $\frac{275}{1292}$ | \frac{275}{1292} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
556fdc78244e131a | 2a1ae2dc844d3766 | tutor | Write a complete step-by-step solution and conclude with the final answer. | How many solutions in integers with $x_i\ge 2$ satisfy $x_1+\cdots+x_9=43$? | 1) Let y_i=x_i-2, then y_i≥0 and y_1+…+y_9=25.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=25 and k=9.
4) Compute C(33,8) = 13884156.
Final answer: $13884156$ | 13884156 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
60588b03b3be929f | 5baffeeac5a4af65 | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Differentiate with respect to x:
$f(x)=- x^{2} - 4$.
Student solution:
1) Differentiate term-by-term.
2) Claim: one term $\frac{d}{dx}(- x^{2}) = - 2 x^{2}$.
3) So $f'(x) = - 2 x^{2}$. | First incorrect step: 2
The error is differentiating a power: the exponent must decrease by 1.
Correct: $\frac{d}{dx}(- x^{2}) = - 2 x$.
Therefore the correct derivative is $f'(x) = - 2 x$.
Final answer: $- 2 x$ | - 2 x | {
"format": "symbolic",
"domain": "calculus",
"topic": "derivative",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
f20069e12120615d | 53c996d8859059ee | verifier | Solve and add a check (substitution/count/sanity check). Provide the final answer. | An urn has 25 red balls and 23 blue balls. If 8 balls are drawn without replacement, what is the probability that exactly 4 are red? | Use a hypergeometric counting argument.
Total equally likely samples: C(48,8).
Favorable samples: choose k reds and n−k blues: C(25,4)·C(23,4).
Compute totals: C(48,8) = 377348994, and favorable = 112015750.
Divide favorable by total and simplify.
Check: computed probability simplifies to $\frac{221375}{745749}$.
Final... | \frac{221375}{745749} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
583429959354af03 | 5271d56e18843f66 | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Differentiate with respect to x:
$f(x)=- 4 x^{2} - 4 x - 2$.
Student solution:
1) Differentiate term-by-term.
2) Claim: one term $\frac{d}{dx}(- 4 x) = - 4 x$.
3) So $f'(x) = - 12 x$. | First incorrect step: 2
The error is differentiating a power: the exponent must decrease by 1.
Correct: $\frac{d}{dx}(- 4 x) = -4$.
Therefore the correct derivative is $f'(x) = 4 \left(- 2 x - 1\right)$.
Final answer: $4 \left(- 2 x - 1\right)$ | 4 \left(- 2 x - 1\right) | {
"format": "symbolic",
"domain": "calculus",
"topic": "derivative",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
df551036224de0e1 | 55df449ba7f9a447 | tutor | Solve carefully and do not skip algebra/arithmetic steps. End with the final answer. | Solve the quadratic equation over the real numbers:
$2 x^{2} - \frac{16 x}{5} - \frac{42}{5}$ = 0. | 1) Identify coefficients a, b, c in ax^2+bx+c=0.
2) Here a=$2$, b=$- \frac{16}{5}$, c=$- \frac{42}{5}$.
3) Compute the discriminant Δ=b^2−4ac.
4) Δ = $\frac{1936}{25}$, so √Δ = $\frac{44}{5}$.
5) Apply the quadratic formula $x=\frac{-b\pm\sqrt{\Delta}}{2a}$.
6) x1 = $3$.
7) x2 = $- \frac{7}{5}$.
8) Collect the solution... | x \in \{3, - \frac{7}{5}\} | {
"format": "symbolic",
"domain": "algebra",
"topic": "quadratic_equation",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
af4aa422324aaf70 | 55df449ba7f9a447 | verifier | Solve step-by-step and include a verification (e.g., plug-in/check totals). Provide the final answer. | Solve the quadratic equation over the real numbers:
$2 x^{2} - \frac{16 x}{5} - \frac{42}{5}$ = 0. | Identify coefficients a, b, c in ax^2+bx+c=0.
Here a=$2$, b=$- \frac{16}{5}$, c=$- \frac{42}{5}$.
Compute the discriminant Δ=b^2−4ac.
Δ = $\frac{1936}{25}$, so √Δ = $\frac{44}{5}$.
Apply the quadratic formula $x=\frac{-b\pm\sqrt{\Delta}}{2a}$.
x1 = $3$.
x2 = $- \frac{7}{5}$.
Collect the solutions into a set.
Check: sub... | x \in \{3, - \frac{7}{5}\} | {
"format": "symbolic",
"domain": "algebra",
"topic": "quadratic_equation",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
45be89b4e3f5c032 | db881649275da537 | concise | Compute and present the key steps only, then the final answer. | Solve the congruences: x≡6 (mod 7), x≡8 (mod 10). Give the solution modulo 70. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 70.
Compute the CRT solution to get x ≡ 48 (mod 70).
Final answer: $x\equiv 48\pmod{70}$ | x\equiv 48\pmod{70} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
fa3a319778421b2f | db881649275da537 | tutor | Solve like a tutor: derive step-by-step and finish with the final answer. | Solve the congruences: x≡6 (mod 7), x≡8 (mod 10). Give the solution modulo 70. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 70.
3) Compute the CRT solution to get x ≡ 48 (mod 70).
Final answer: $x\equiv 48\pmod{70}$ | x\equiv 48\pmod{70} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
a589f6bae048c6ec | 80b56951684bb166 | tutor | Show a full solution with explicit computations and a final answer line. | An urn has 24 red balls and 23 blue balls. If 8 balls are drawn without replacement, what is the probability that exactly 2 are red? | 1) Use a hypergeometric counting argument.
2) Total equally likely samples: C(47,8).
3) Favorable samples: choose k reds and n−k blues: C(24,2)·C(23,6).
4) Compute totals: C(47,8) = 314457495, and favorable = 27861372.
5) Divide favorable by total and simplify.
Final answer: $\frac{36708}{414305}$ | \frac{36708}{414305} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
2fcf1ecfa3b2b12f | 80b56951684bb166 | verifier | Solve and add a check (substitution/count/sanity check). Provide the final answer. | An urn has 24 red balls and 23 blue balls. If 8 balls are drawn without replacement, what is the probability that exactly 2 are red? | Use a hypergeometric counting argument.
Total equally likely samples: C(47,8).
Favorable samples: choose k reds and n−k blues: C(24,2)·C(23,6).
Compute totals: C(47,8) = 314457495, and favorable = 27861372.
Divide favorable by total and simplify.
Check: computed probability simplifies to $\frac{36708}{414305}$.
Final a... | \frac{36708}{414305} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
01b437df633d47b4 | 354c9edb8bba55e3 | tutor | Provide a thorough solution as if teaching, including intermediate computations, then give the final answer. | A coin lands heads with probability p=$\frac{1}{4}$. If it is flipped 7 times, what is the probability of exactly 2 heads? | 1) Let X be the number of heads in n independent flips. Then X ~ Bin(n,p).
2) Use $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
3) Substitute n=7, k=2, p=$\frac{1}{4}$ and simplify.
4) Here C(7,2) = 21.
Final answer: $\frac{5103}{16384}$ | \frac{5103}{16384} | {
"format": "symbolic",
"domain": "probability",
"topic": "binomial_probability",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
47070b67c331a32f | 354c9edb8bba55e3 | verifier | Solve and add a check (substitution/count/sanity check). Provide the final answer. | A coin lands heads with probability p=$\frac{1}{4}$. If it is flipped 7 times, what is the probability of exactly 2 heads? | Let X be the number of heads in n independent flips. Then X ~ Bin(n,p).
Use $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
Substitute n=7, k=2, p=$\frac{1}{4}$ and simplify.
Here C(7,2) = 21.
Check: expression simplifies to $\frac{5103}{16384}$.
Final answer: $\frac{5103}{16384}$ | \frac{5103}{16384} | {
"format": "symbolic",
"domain": "probability",
"topic": "binomial_probability",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
5330a76ab9aa279d | f3fbe5f75fc6bb02 | tutor | Provide a detailed solution with numbered steps and a final answer line. | An urn has 25 red balls and 14 blue balls. If 11 balls are drawn without replacement, what is the probability that exactly 2 are red? | 1) Use a hypergeometric counting argument.
2) Total equally likely samples: C(39,11).
3) Favorable samples: choose k reds and n−k blues: C(25,2)·C(14,9).
4) Compute totals: C(39,11) = 1676056044, and favorable = 600600.
5) Divide favorable by total and simplify.
Final answer: $\frac{3850}{10743949}$ | \frac{3850}{10743949} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
8902239c0e20a21d | f3fbe5f75fc6bb02 | answer_only | Provide only the final answer (no steps). | An urn has 25 red balls and 14 blue balls. If 11 balls are drawn without replacement, what is the probability that exactly 2 are red? | $\frac{3850}{10743949}$ | \frac{3850}{10743949} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
19dfba3cf6b38ba3 | fbd5cd6ebb3e228b | tutor | Give a detailed derivation, showing intermediate expressions, and then the final answer. | An urn has 13 red balls and 9 blue balls. If 6 balls are drawn without replacement, what is the probability that exactly 3 are red? | 1) Use a hypergeometric counting argument.
2) Total equally likely samples: C(22,6).
3) Favorable samples: choose k reds and n−k blues: C(13,3)·C(9,3).
4) Compute totals: C(22,6) = 74613, and favorable = 24024.
5) Divide favorable by total and simplify.
Final answer: $\frac{104}{323}$ | \frac{104}{323} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
ef784187eeeef03f | fbd5cd6ebb3e228b | verifier | Compute the answer and add a brief verification calculation. Provide the final answer. | An urn has 13 red balls and 9 blue balls. If 6 balls are drawn without replacement, what is the probability that exactly 3 are red? | Use a hypergeometric counting argument.
Total equally likely samples: C(22,6).
Favorable samples: choose k reds and n−k blues: C(13,3)·C(9,3).
Compute totals: C(22,6) = 74613, and favorable = 24024.
Divide favorable by total and simplify.
Check: computed probability simplifies to $\frac{104}{323}$.
Final answer: $\frac... | \frac{104}{323} | {
"format": "symbolic",
"domain": "probability",
"topic": "hypergeometric",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
fb700dfef2344ace | ef8b18d9cfe420e7 | tutor | Provide a thorough solution as if teaching, including intermediate computations, then give the final answer. | Find the multiplicative inverse of 12 modulo 31. | 1) We want x such that 12x ≡ 1 (mod 31).
2) Use extended Euclid to find s,t with 12s + 31t = 1.
3) One Bezout identity is: 12·(13) + 31·(-5) = 1.
4) Reduce s modulo 31: x ≡ 13 (mod 31).
Final answer: $x\equiv 13\pmod{31}$ | x\equiv 13\pmod{31} | {
"format": "symbolic",
"domain": "discrete",
"topic": "modular_inverse",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
d68a959fd3c9d57f | d3b067b3b14a6674 | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Differentiate with respect to x:
$f(x)=- 3 x^{3} \sin{\left(x \right)} + 2 x \sin{\left(x \right)} + \sin{\left(x \right)} + 3 \cos{\left(2 x \right)}$.
Student solution:
1) Differentiate term-by-term.
2) Claim (missing chain rule): $\frac{d}{dx}(2 x \sin{\left(x \right)}) = 2 x \cos{\left(x \right)}$.
3) So $f'(x) = ... | First incorrect step: 2
The error is omitting the chain-rule factor (derivative of the inside function).
Correct derivative of that term is $\frac{d}{dx}(2 x \sin{\left(x \right)}) = 2 x \cos{\left(x \right)} + 2 \sin{\left(x \right)}$.
Therefore the correct derivative is $f'(x) = - 3 x^{3} \cos{\left(x \right)} - 9 x^... | - 3 x^{3} \cos{\left(x \right)} - 9 x^{2} \sin{\left(x \right)} + 2 x \cos{\left(x \right)} + 2 \sin{\left(x \right)} - 6 \sin{\left(2 x \right)} + \cos{\left(x \right)} | {
"format": "symbolic",
"domain": "calculus",
"topic": "derivative",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
d3a338f673515d08 | b39c1a91a816552e | concise | Solve and give a compact solution, then the final answer. | Solve the congruences: x≡3 (mod 11), x≡7 (mod 13). Give the solution modulo 143. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 143.
Compute the CRT solution to get x ≡ 124 (mod 143).
Final answer: $x\equiv 124\pmod{143}$ | x\equiv 124\pmod{143} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
9e24bfdbe89985c1 | b39c1a91a816552e | verifier | Compute the answer and add a brief verification calculation. Provide the final answer. | Solve the congruences: x≡3 (mod 11), x≡7 (mod 13). Give the solution modulo 143. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 143.
Compute the CRT solution to get x ≡ 124 (mod 143).
Check: 124 mod 11=3 and mod 13=7.
Final answer: $x\equiv 124\pmod{143}$ | x\equiv 124\pmod{143} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
a0545474f6e88a58 | 989df28cc09af7b6 | tutor | Explain your reasoning clearly and show all important intermediate computations. End with the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_4=16$? | 1) We are counting nonnegative integer solutions to a sum constraint.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=16 and k=4.
4) Compute C(19,3) = 969.
Final answer: $969$ | 969 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
55cc091e4cf6907e | 989df28cc09af7b6 | verifier | Compute the answer and add a brief verification calculation. Provide the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_4=16$? | We are counting nonnegative integer solutions to a sum constraint.
Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
Here N=16 and k=4.
Compute C(19,3) = 969.
Check: matches the standard stars-and-bars formula.
Final answer: $969$ | 969 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
bbb586c445d455f1 | 34f54e5ed7a5834a | tutor | Work through the problem carefully with explicit steps and intermediate values. Finish with the final answer. | Solve the congruences: x≡0 (mod 3), x≡5 (mod 7). Give the solution modulo 21. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 21.
3) Compute the CRT solution to get x ≡ 12 (mod 21).
Final answer: $x\equiv 12\pmod{21}$ | x\equiv 12\pmod{21} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
a651593380a05353 | 34f54e5ed7a5834a | verifier | Solve the problem, then verify the result with a quick check. Provide the final answer. | Solve the congruences: x≡0 (mod 3), x≡5 (mod 7). Give the solution modulo 21. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 21.
Compute the CRT solution to get x ≡ 12 (mod 21).
Check: 12 mod 3=0 and mod 7=5.
Final answer: $x\equiv 12\pmod{21}$ | x\equiv 12\pmod{21} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
e1fd938c8004754d | 8bb6c802a0bd2cdf | tutor | Show your work in clear steps (with intermediate values) and finish with a final answer. | Solve the congruences: x≡1 (mod 5), x≡0 (mod 11). Give the solution modulo 55. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 55.
3) Compute the CRT solution to get x ≡ 11 (mod 55).
Final answer: $x\equiv 11\pmod{55}$ | x\equiv 11\pmod{55} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
5d7bebd0eb5e0f77 | 5377178775cdd578 | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Differentiate with respect to x:
$f(x)=- x^{2} - 4 x$.
Student solution:
1) Differentiate term-by-term.
2) Claim: one term $\frac{d}{dx}(- x^{2}) = - 2 x^{2}$.
3) So $f'(x) = - 2 x^{2} - 4$. | First incorrect step: 2
The error is differentiating a power: the exponent must decrease by 1.
Correct: $\frac{d}{dx}(- x^{2}) = - 2 x$.
Therefore the correct derivative is $f'(x) = 2 \left(- x - 2\right)$.
Final answer: $2 \left(- x - 2\right)$ | 2 \left(- x - 2\right) | {
"format": "symbolic",
"domain": "calculus",
"topic": "derivative",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
cf9b324192e833ce | 90c50fcb74a6ea49 | tutor | Solve step-by-step like a math tutor. Show intermediate calculations and end with a final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_7=43$? | 1) We are counting nonnegative integer solutions to a sum constraint.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=43 and k=7.
4) Compute C(49,6) = 13983816.
Final answer: $13983816$ | 13983816 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
475cc78c267d0d72 | 9d4cd9c05c99030c | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Evaluate the limit:
$\displaystyle \lim_{x\to 2} x^{3} + 4 x^{2} + 7 x + 13$.
Student solution:
1) Substitute x=2 into the expression.
2) (Mistake) Simplify the expression to x^{3} + 4 x^{2} + 6 x + 13.
3) Then the limit is 49. | First incorrect step: 2
The error is dropping/altering terms before taking the limit.
Use the original expression and substitute x=2 (polynomial is continuous).
Correct limit value: 51.
Final answer: $51$ | 51 | {
"format": "symbolic",
"domain": "calculus",
"topic": "limit",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
b08f1e64150350e3 | 0e76c1b819c8c9ff | tutor | Solve like a tutor: derive step-by-step and finish with the final answer. | Solve the congruences: x≡7 (mod 11), x≡0 (mod 5). Give the solution modulo 55. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 55.
3) Compute the CRT solution to get x ≡ 40 (mod 55).
Final answer: $x\equiv 40\pmod{55}$ | x\equiv 40\pmod{55} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
d5a88b52e2adf857 | 0e76c1b819c8c9ff | verifier | Solve step-by-step and include a verification (e.g., plug-in/check totals). Provide the final answer. | Solve the congruences: x≡7 (mod 11), x≡0 (mod 5). Give the solution modulo 55. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 55.
Compute the CRT solution to get x ≡ 40 (mod 55).
Check: 40 mod 11=7 and mod 5=0.
Final answer: $x\equiv 40\pmod{55}$ | x\equiv 40\pmod{55} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
b0259f5aa1c158dd | 7e8ae5a1ffa3976d | concise | Provide a brief derivation and the final answer. | A coin lands heads with probability p=$\frac{3}{10}$. If it is flipped 9 times, what is the probability of exactly 7 heads? | Let X be the number of heads in n independent flips. Then X ~ Bin(n,p).
Use $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
Substitute n=9, k=7, p=$\frac{3}{10}$ and simplify.
Final answer: $\frac{964467}{250000000}$ | \frac{964467}{250000000} | {
"format": "symbolic",
"domain": "probability",
"topic": "binomial_probability",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
3b4e17f9be4b4a0d | 7e8ae5a1ffa3976d | verifier | Solve step-by-step and include a verification (e.g., plug-in/check totals). Provide the final answer. | A coin lands heads with probability p=$\frac{3}{10}$. If it is flipped 9 times, what is the probability of exactly 7 heads? | Let X be the number of heads in n independent flips. Then X ~ Bin(n,p).
Use $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
Substitute n=9, k=7, p=$\frac{3}{10}$ and simplify.
Here C(9,7) = 36.
Check: expression simplifies to $\frac{964467}{250000000}$.
Final answer: $\frac{964467}{250000000}$ | \frac{964467}{250000000} | {
"format": "symbolic",
"domain": "probability",
"topic": "binomial_probability",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
874a852da4c09ac2 | 230da4c41ab57047 | tutor | Solve carefully and do not skip algebra/arithmetic steps. End with the final answer. | Solve the congruences: x≡5 (mod 11), x≡0 (mod 13). Give the solution modulo 143. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 143.
3) Compute the CRT solution to get x ≡ 104 (mod 143).
Final answer: $x\equiv 104\pmod{143}$ | x\equiv 104\pmod{143} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
4e025f3483e75dd5 | 230da4c41ab57047 | verifier | Compute the answer and add a brief verification calculation. Provide the final answer. | Solve the congruences: x≡5 (mod 11), x≡0 (mod 13). Give the solution modulo 143. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 143.
Compute the CRT solution to get x ≡ 104 (mod 143).
Check: 104 mod 11=5 and mod 13=0.
Final answer: $x\equiv 104\pmod{143}$ | x\equiv 104\pmod{143} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
f74f600273dab56e | 46c842cba097b97b | tutor | Solve carefully and do not skip algebra/arithmetic steps. End with the final answer. | Solve the congruences: x≡3 (mod 4), x≡1 (mod 9). Give the solution modulo 36. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 36.
3) Compute the CRT solution to get x ≡ 19 (mod 36).
Final answer: $x\equiv 19\pmod{36}$ | x\equiv 19\pmod{36} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
ffa6cfec71bcd36e | 46c842cba097b97b | verifier | Compute the answer and add a brief verification calculation. Provide the final answer. | Solve the congruences: x≡3 (mod 4), x≡1 (mod 9). Give the solution modulo 36. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 36.
Compute the CRT solution to get x ≡ 19 (mod 36).
Check: 19 mod 4=3 and mod 9=1.
Final answer: $x\equiv 19\pmod{36}$ | x\equiv 19\pmod{36} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
6cf606d7395ff53a | a03cecb587f179af | tutor | Explain your reasoning clearly and show all important intermediate computations. End with the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_6=13$? | 1) We are counting nonnegative integer solutions to a sum constraint.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=13 and k=6.
4) Compute C(18,5) = 8568.
Final answer: $8568$ | 8568 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
f3b6697b20b77372 | a03cecb587f179af | verifier | Show the solution, then validate it with a quick check. End with the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_6=13$? | We are counting nonnegative integer solutions to a sum constraint.
Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
Here N=13 and k=6.
Compute C(18,5) = 8568.
Check: matches the standard stars-and-bars formula.
Final answer: $8568$ | 8568 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
d279fcab7547e212 | 0d4699c29e15c8be | tutor | Solve step-by-step like a math tutor. Show intermediate calculations and end with a final answer. | Solve the congruences: x≡3 (mod 4), x≡0 (mod 7). Give the solution modulo 28. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 28.
3) Compute the CRT solution to get x ≡ 7 (mod 28).
Final answer: $x\equiv 7\pmod{28}$ | x\equiv 7\pmod{28} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
e53347a5420acddd | e1636336dac3aceb | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Differentiate with respect to x:
$f(x)=2 x^{4} - 2 x^{3} + 2 x^{2} + 4 x + 2$.
Student solution:
1) Differentiate term-by-term.
2) Claim: one term $\frac{d}{dx}(- 2 x^{3}) = - 6 x^{3}$.
3) So $f'(x) = 2 x^{3} + 4 x + 4$. | First incorrect step: 2
The error is differentiating a power: the exponent must decrease by 1.
Correct: $\frac{d}{dx}(- 2 x^{3}) = - 6 x^{2}$.
Therefore the correct derivative is $f'(x) = 2 \left(4 x^{3} - 3 x^{2} + 2 x + 2\right)$.
Final answer: $2 \left(4 x^{3} - 3 x^{2} + 2 x + 2\right)$ | 2 \left(4 x^{3} - 3 x^{2} + 2 x + 2\right) | {
"format": "symbolic",
"domain": "calculus",
"topic": "derivative",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
d4e9f019f9bc5cc6 | 5948e360f6cfcc01 | tutor | Solve carefully and do not skip algebra/arithmetic steps. End with the final answer. | Solve the congruences: x≡3 (mod 5), x≡4 (mod 9). Give the solution modulo 45. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 45.
3) Compute the CRT solution to get x ≡ 13 (mod 45).
Final answer: $x\equiv 13\pmod{45}$ | x\equiv 13\pmod{45} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
4963720c1fed8ead | 5948e360f6cfcc01 | verifier | Solve step-by-step and include a verification (e.g., plug-in/check totals). Provide the final answer. | Solve the congruences: x≡3 (mod 5), x≡4 (mod 9). Give the solution modulo 45. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 45.
Compute the CRT solution to get x ≡ 13 (mod 45).
Check: 13 mod 5=3 and mod 9=4.
Final answer: $x\equiv 13\pmod{45}$ | x\equiv 13\pmod{45} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
71ad8a9b3cd99352 | 07a26bdf3d31706b | tutor | Give a detailed derivation, showing intermediate expressions, and then the final answer. | Find the multiplicative inverse of 6 modulo 29. | 1) We want x such that 6x ≡ 1 (mod 29).
2) Use extended Euclid to find s,t with 6s + 29t = 1.
3) One Bezout identity is: 6·(5) + 29·(-1) = 1.
4) Reduce s modulo 29: x ≡ 5 (mod 29).
Final answer: $x\equiv 5\pmod{29}$ | x\equiv 5\pmod{29} | {
"format": "symbolic",
"domain": "discrete",
"topic": "modular_inverse",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
3fd23abc27c872db | 07a26bdf3d31706b | verifier | Solve and add a check (substitution/count/sanity check). Provide the final answer. | Find the multiplicative inverse of 6 modulo 29. | We want x such that 6x ≡ 1 (mod 29).
Use extended Euclid to find s,t with 6s + 29t = 1.
One Bezout identity is: 6·(5) + 29·(-1) = 1.
Reduce s modulo 29: x ≡ 5 (mod 29).
Check: 6·5 mod 29 = 1.
Final answer: $x\equiv 5\pmod{29}$ | x\equiv 5\pmod{29} | {
"format": "symbolic",
"domain": "discrete",
"topic": "modular_inverse",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
471220ec90b1c5c9 | dc4a53249735b962 | verifier | Solve and add a check (substitution/count/sanity check). Provide the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_8=41$? | We are counting nonnegative integer solutions to a sum constraint.
Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
Here N=41 and k=8.
Compute C(48,7) = 73629072.
Check: matches the standard stars-and-bars formula.
Final answer: $73629072$ | 73629072 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
2053f4f46064d47c | dc4a53249735b962 | tutor | Show a full solution with explicit computations and a final answer line. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_8=41$? | 1) We are counting nonnegative integer solutions to a sum constraint.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=41 and k=8.
4) Compute C(48,7) = 73629072.
Final answer: $73629072$ | 73629072 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
dfabd15101e0dbc8 | 796dda9e4de67fe2 | tutor | Explain your reasoning clearly and show all important intermediate computations. End with the final answer. | Find the multiplicative inverse of 24 modulo 53. | 1) We want x such that 24x ≡ 1 (mod 53).
2) Use extended Euclid to find s,t with 24s + 53t = 1.
3) One Bezout identity is: 24·(-11) + 53·(5) = 1.
4) Reduce s modulo 53: x ≡ 42 (mod 53).
Final answer: $x\equiv 42\pmod{53}$ | x\equiv 42\pmod{53} | {
"format": "symbolic",
"domain": "discrete",
"topic": "modular_inverse",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
b9919886fc13f190 | 796dda9e4de67fe2 | verifier | Show the solution, then validate it with a quick check. End with the final answer. | Find the multiplicative inverse of 24 modulo 53. | We want x such that 24x ≡ 1 (mod 53).
Use extended Euclid to find s,t with 24s + 53t = 1.
One Bezout identity is: 24·(-11) + 53·(5) = 1.
Reduce s modulo 53: x ≡ 42 (mod 53).
Check: 24·42 mod 53 = 1.
Final answer: $x\equiv 42\pmod{53}$ | x\equiv 42\pmod{53} | {
"format": "symbolic",
"domain": "discrete",
"topic": "modular_inverse",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
c4a468e7fa244e12 | 0f668de8c40d7715 | tutor | Solve like a tutor: derive step-by-step and finish with the final answer. | Solve the congruences: x≡3 (mod 5), x≡2 (mod 11). Give the solution modulo 55. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 55.
3) Compute the CRT solution to get x ≡ 13 (mod 55).
Final answer: $x\equiv 13\pmod{55}$ | x\equiv 13\pmod{55} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
f067218191dda17a | 0f668de8c40d7715 | verifier | Solve and add a check (substitution/count/sanity check). Provide the final answer. | Solve the congruences: x≡3 (mod 5), x≡2 (mod 11). Give the solution modulo 55. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 55.
Compute the CRT solution to get x ≡ 13 (mod 55).
Check: 13 mod 5=3 and mod 11=2.
Final answer: $x\equiv 13\pmod{55}$ | x\equiv 13\pmod{55} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
1663cbbe08cede96 | fea9e13c4d62a1b4 | tutor | Provide a thorough solution as if teaching, including intermediate computations, then give the final answer. | Solve the congruences: x≡0 (mod 11), x≡3 (mod 9). Give the solution modulo 99. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 99.
3) Compute the CRT solution to get x ≡ 66 (mod 99).
Final answer: $x\equiv 66\pmod{99}$ | x\equiv 66\pmod{99} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
d908e64ee75bc185 | fea9e13c4d62a1b4 | verifier | Show the solution, then validate it with a quick check. End with the final answer. | Solve the congruences: x≡0 (mod 11), x≡3 (mod 9). Give the solution modulo 99. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 99.
Compute the CRT solution to get x ≡ 66 (mod 99).
Check: 66 mod 11=0 and mod 9=3.
Final answer: $x\equiv 66\pmod{99}$ | x\equiv 66\pmod{99} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
cc815ec26695f575 | 67f008ad6d8fa665 | tutor | Show a full solution with explicit computations and a final answer line. | Solve the congruences: x≡0 (mod 8), x≡4 (mod 5). Give the solution modulo 40. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 40.
3) Compute the CRT solution to get x ≡ 24 (mod 40).
Final answer: $x\equiv 24\pmod{40}$ | x\equiv 24\pmod{40} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
8fafd4f614ab175c | 67f008ad6d8fa665 | concise | Solve the problem and keep the explanation concise but complete. Provide the final answer. | Solve the congruences: x≡0 (mod 8), x≡4 (mod 5). Give the solution modulo 40. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 40.
Compute the CRT solution to get x ≡ 24 (mod 40).
Final answer: $x\equiv 24\pmod{40}$ | x\equiv 24\pmod{40} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
74a84809fd9bd220 | 19f5bee37889f6af | tutor | Give a detailed derivation, showing intermediate expressions, and then the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_6=39$? | 1) We are counting nonnegative integer solutions to a sum constraint.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=39 and k=6.
4) Compute C(44,5) = 1086008.
Final answer: $1086008$ | 1086008 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
fa326f848618e8cd | 071fad801159a5a8 | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Evaluate the limit:
$\displaystyle \lim_{x\to 1} x^{3} - x^{2} - 2 x - 5$.
Student solution:
1) Substitute x=1 into the expression.
2) (Mistake) Simplify the expression to x^{3} - x^{2} - 3 x - 5.
3) Then the limit is -8. | First incorrect step: 2
The error is dropping/altering terms before taking the limit.
Use the original expression and substitute x=1 (polynomial is continuous).
Correct limit value: -7.
Final answer: $-7$ | -7 | {
"format": "symbolic",
"domain": "calculus",
"topic": "limit",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
80352f8c51947f3b | 983be6f721f254ba | verifier | Compute the answer and add a brief verification calculation. Provide the final answer. | Solve the congruences: x≡3 (mod 5), x≡4 (mod 13). Give the solution modulo 65. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 65.
Compute the CRT solution to get x ≡ 43 (mod 65).
Check: 43 mod 5=3 and mod 13=4.
Final answer: $x\equiv 43\pmod{65}$ | x\equiv 43\pmod{65} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
bd88f0b84533b42e | 983be6f721f254ba | tutor | Write a complete step-by-step solution and conclude with the final answer. | Solve the congruences: x≡3 (mod 5), x≡4 (mod 13). Give the solution modulo 65. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 65.
3) Compute the CRT solution to get x ≡ 43 (mod 65).
Final answer: $x\equiv 43\pmod{65}$ | x\equiv 43\pmod{65} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
e1e7e9a2e68cfae9 | 7e99da9d5364a799 | tutor | Solve like a tutor: derive step-by-step and finish with the final answer. | Solve the congruences: x≡5 (mod 7), x≡1 (mod 9). Give the solution modulo 63. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 63.
3) Compute the CRT solution to get x ≡ 19 (mod 63).
Final answer: $x\equiv 19\pmod{63}$ | x\equiv 19\pmod{63} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
239329f1f4413589 | 7e99da9d5364a799 | verifier | Compute the answer and include a short verification step. Give the final answer. | Solve the congruences: x≡5 (mod 7), x≡1 (mod 9). Give the solution modulo 63. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 63.
Compute the CRT solution to get x ≡ 19 (mod 63).
Check: 19 mod 7=5 and mod 9=1.
Final answer: $x\equiv 19\pmod{63}$ | x\equiv 19\pmod{63} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 1,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
8cf6c140d20904c3 | 3b104a320d2c4320 | tutor | Provide a detailed solution with numbered steps and a final answer line. | How many solutions in integers with $x_i\ge 1$ satisfy $x_1+\cdots+x_10=35$? | 1) Let y_i=x_i-1, then y_i≥0 and y_1+…+y_10=25.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=25 and k=10.
4) Compute C(34,9) = 52451256.
Final answer: $52451256$ | 52451256 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
1fee33264e95a68c | 3b104a320d2c4320 | concise | Compute and present the key steps only, then the final answer. | How many solutions in integers with $x_i\ge 1$ satisfy $x_1+\cdots+x_10=35$? | Let y_i=x_i-1, then y_i≥0 and y_1+…+y_10=25.
Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
Here N=25 and k=10.
Final answer: $52451256$ | 52451256 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 4,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
b664b6bfbe21b17c | 23093d2102e6fe2a | tutor | Explain your reasoning clearly and show all important intermediate computations. End with the final answer. | Solve the congruences: x≡3 (mod 8), x≡10 (mod 11). Give the solution modulo 88. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 88.
3) Compute the CRT solution to get x ≡ 43 (mod 88).
Final answer: $x\equiv 43\pmod{88}$ | x\equiv 43\pmod{88} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
d153d60f8636360b | 23093d2102e6fe2a | verifier | Show the solution, then validate it with a quick check. End with the final answer. | Solve the congruences: x≡3 (mod 8), x≡10 (mod 11). Give the solution modulo 88. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 88.
Compute the CRT solution to get x ≡ 43 (mod 88).
Check: 43 mod 8=3 and mod 11=10.
Final answer: $x\equiv 43\pmod{88}$ | x\equiv 43\pmod{88} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
21ac11dbd29abf1b | bb58e7dd0582f569 | tutor | Solve like a tutor: derive step-by-step and finish with the final answer. | Solve the congruences: x≡2 (mod 7), x≡7 (mod 11). Give the solution modulo 77. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 77.
3) Compute the CRT solution to get x ≡ 51 (mod 77).
Final answer: $x\equiv 51\pmod{77}$ | x\equiv 51\pmod{77} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
65e277eb88f33994 | bb58e7dd0582f569 | concise | Solve with minimal but sufficient steps, then provide the final answer. | Solve the congruences: x≡2 (mod 7), x≡7 (mod 11). Give the solution modulo 77. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 77.
Compute the CRT solution to get x ≡ 51 (mod 77).
Final answer: $x\equiv 51\pmod{77}$ | x\equiv 51\pmod{77} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
ace8515e46f06547 | bdea911768e94544 | tutor | Solve carefully and do not skip algebra/arithmetic steps. End with the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_3=15$? | 1) We are counting nonnegative integer solutions to a sum constraint.
2) Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
3) Here N=15 and k=3.
4) Compute C(17,2) = 136.
Final answer: $136$ | 136 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
6864c05ceb10d0b4 | bdea911768e94544 | concise | Give a short solution with the key steps and the final answer. | How many solutions in nonnegative integers satisfy $x_1+\cdots+x_3=15$? | We are counting nonnegative integer solutions to a sum constraint.
Stars and bars: the number of solutions to y1+…+yk=N with yi≥0 is C(N+k−1,k−1).
Here N=15 and k=3.
Final answer: $136$ | 136 | {
"format": "symbolic",
"domain": "discrete",
"topic": "stars_and_bars",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
922b41ea900def15 | 5928c43e7508f11f | concise | Compute and present the key steps only, then the final answer. | Solve the congruences: x≡0 (mod 8), x≡1 (mod 5). Give the solution modulo 40. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 40.
Compute the CRT solution to get x ≡ 16 (mod 40).
Final answer: $x\equiv 16\pmod{40}$ | x\equiv 16\pmod{40} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
1c2ce2de894dee2e | 5928c43e7508f11f | tutor | Work through the problem carefully with explicit steps and intermediate values. Finish with the final answer. | Solve the congruences: x≡0 (mod 8), x≡1 (mod 5). Give the solution modulo 40. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 40.
3) Compute the CRT solution to get x ≡ 16 (mod 40).
Final answer: $x\equiv 16\pmod{40}$ | x\equiv 16\pmod{40} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
af049e0caeb483b5 | f4b7ec4f3e55ef27 | tutor | Provide a thorough solution as if teaching, including intermediate computations, then give the final answer. | Solve the congruences: x≡2 (mod 8), x≡11 (mod 13). Give the solution modulo 104. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 104.
3) Compute the CRT solution to get x ≡ 50 (mod 104).
Final answer: $x\equiv 50\pmod{104}$ | x\equiv 50\pmod{104} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
1f0d518ee42c73f7 | f4b7ec4f3e55ef27 | verifier | Solve and add a check (substitution/count/sanity check). Provide the final answer. | Solve the congruences: x≡2 (mod 8), x≡11 (mod 13). Give the solution modulo 104. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 104.
Compute the CRT solution to get x ≡ 50 (mod 104).
Check: 50 mod 8=2 and mod 13=11.
Final answer: $x\equiv 50\pmod{104}$ | x\equiv 50\pmod{104} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
5336bcd494a9e58a | 12717b20a9259407 | verifier_negative | A student attempted the solution below. Identify the FIRST incorrect step, explain the mistake precisely, then give the corrected solution and final answer. | Differentiate with respect to x:
$f(x)=2 x^{3} \sin{\left(x \right)} - 2 x^{2} \sin{\left(x \right)} + x \sin{\left(x \right)} + 2 \sin{\left(x \right)} + \cos{\left(2 x \right)}$.
Student solution:
1) Differentiate term-by-term.
2) Claim (missing chain rule): $\frac{d}{dx}(2 x^{3} \sin{\left(x \right)}) = 2 x^{3} \co... | First incorrect step: 2
The error is omitting the chain-rule factor (derivative of the inside function).
Correct derivative of that term is $\frac{d}{dx}(2 x^{3} \sin{\left(x \right)}) = 2 x^{3} \cos{\left(x \right)} + 6 x^{2} \sin{\left(x \right)}$.
Therefore the correct derivative is $f'(x) = 2 x^{3} \cos{\left(x \ri... | 2 x^{3} \cos{\left(x \right)} + 6 x^{2} \sin{\left(x \right)} - 2 x^{2} \cos{\left(x \right)} - 4 x \sin{\left(x \right)} + x \cos{\left(x \right)} + \sin{\left(x \right)} - 2 \sin{\left(2 x \right)} + 2 \cos{\left(x \right)} | {
"format": "symbolic",
"domain": "calculus",
"topic": "derivative",
"difficulty": 3,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0",
"task": "error_localization",
"is_negative": true
} |
dcdc4fb9569dd8e1 | 3d472c36dd2badf6 | tutor | Explain each step clearly and end with the final answer. | Solve the congruences: x≡8 (mod 9), x≡4 (mod 10). Give the solution modulo 90. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 90.
3) Compute the CRT solution to get x ≡ 44 (mod 90).
Final answer: $x\equiv 44\pmod{90}$ | x\equiv 44\pmod{90} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
c8c81f9a69aef14c | 7255aafecc2c075c | tutor | Solve step-by-step like a math tutor. Show intermediate calculations and end with a final answer. | Solve the congruences: x≡2 (mod 7), x≡12 (mod 13). Give the solution modulo 91. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 91.
3) Compute the CRT solution to get x ≡ 51 (mod 91).
Final answer: $x\equiv 51\pmod{91}$ | x\equiv 51\pmod{91} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
ed1daf3b4ec67134 | 7255aafecc2c075c | verifier | Solve step-by-step and include a verification (e.g., plug-in/check totals). Provide the final answer. | Solve the congruences: x≡2 (mod 7), x≡12 (mod 13). Give the solution modulo 91. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 91.
Compute the CRT solution to get x ≡ 51 (mod 91).
Check: 51 mod 7=2 and mod 13=12.
Final answer: $x\equiv 51\pmod{91}$ | x\equiv 51\pmod{91} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 5,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
4893980eeacf8455 | 09f5e2c4f465c239 | tutor | Provide a thorough solution as if teaching, including intermediate computations, then give the final answer. | Solve the congruences: x≡4 (mod 9), x≡7 (mod 8). Give the solution modulo 72. | 1) Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
2) The solution is unique modulo m1·m2 = 72.
3) Compute the CRT solution to get x ≡ 31 (mod 72).
Final answer: $x\equiv 31\pmod{72}$ | x\equiv 31\pmod{72} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
d42b88e0ac1b6c8e | 09f5e2c4f465c239 | verifier | Solve and add a check (substitution/count/sanity check). Provide the final answer. | Solve the congruences: x≡4 (mod 9), x≡7 (mod 8). Give the solution modulo 72. | Since gcd(m1,m2)=1, the Chinese Remainder Theorem applies.
The solution is unique modulo m1·m2 = 72.
Compute the CRT solution to get x ≡ 31 (mod 72).
Check: 31 mod 9=4 and mod 8=7.
Final answer: $x\equiv 31\pmod{72}$ | x\equiv 31\pmod{72} | {
"format": "symbolic",
"domain": "discrete",
"topic": "chinese_remainder_theorem",
"difficulty": 2,
"verified": true,
"source": "synthetic_v3_open",
"sympy_version": "1.14.0"
} |
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