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- ==Problem== #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]1 KB (164 words) - 16:53, 11 July 2021
- ==Problem== #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]1 KB (197 words) - 09:01, 12 July 2021
- ==Problem== ...lue of <math>60</math> dimes is: <math>60 \cdot 0.10=6.00</math>, or <math>6</math> dollars.1 KB (175 words) - 16:13, 11 July 2021
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- ==Problem== ...math>252=2^2 \cdot 3^2 \cdot 7</math>, so the divisors are <math>\{1,2,3,4,6,7,9,12,14,18,21,28,36,42,63,84,126,252 \}</math>. We see the set <math>\{212 KB (296 words) - 00:17, 12 July 2021
- ...(Middle School)''' problems and solutions. The test was held on July 10th, 2021. *[[2021 JMPSC Accuracy Problems]]1 KB (125 words) - 16:39, 11 July 2021
- ...Middle School</b> problems and solutions. The test was held on July 10th, 2021. *[[2021 JMPSC Sprint Problems]]1 KB (155 words) - 16:39, 11 July 2021
- == Problem == Compute <math>\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)(3+6+9)</math>.4 KB (634 words) - 09:53, 12 July 2021
- #You will receive 4 points for each correct answer, and 0 points for each problem left unanswered or incorrect. ...rasers. No calculators, smartwatches, or computing devices are allowed. No problems on the test will require the use of a calculator.5 KB (729 words) - 17:41, 11 July 2021
- ==Problem== Say we take <math>x_1,x_1,x_3,...,x_{10}</math> as <math>4,5,6,...,13</math> as an example. The first few terms of the inequality would th1 KB (178 words) - 16:24, 11 July 2021
- ==Problem== If <math>\frac{x+2}{6}</math> is its own reciprocal, find the product of all possible values of <1 KB (205 words) - 16:23, 11 July 2021
- ==Problem== .../math>, so <math>A=2</math> and <math>B=6</math>. The answer is <math>3(2)+6(2)+1(5)=\boxed{23}</math>2 KB (332 words) - 09:42, 18 July 2021
- ==Problem== ...lue of <math>60</math> dimes is: <math>60 \cdot 0.10=6.00</math>, or <math>6</math> dollars.1 KB (175 words) - 16:13, 11 July 2021
- ==Problem== If the machine splits the metal into <math>6</math> pieces, then it made <math>5</math> cuts. Therefore, it makes <math>797 bytes (121 words) - 16:14, 11 July 2021
- ==Problem== ...dot 2}.</cmath> We square both sides of the equation again to get <cmath>x^6 \cdot x=x^7=2^{63 \cdot 4}.</cmath> Thus, <math>x=2^{63 \cdot 4/7}=2^{36}</2 KB (367 words) - 19:02, 12 July 2021
- #You will receive 3 points for each correct answer, and 0 points for each problem left unanswered or incorrect. ...rasers. No calculators, smartwatches, or computing devices are allowed. No problems on the test will require the use of a calculator.7 KB (1,100 words) - 17:40, 11 July 2021
- ...(Middle School)''' problems and solutions. The test was held on July 11th, 2021. *[[2021 JMPSC Invitationals Problems]]1 KB (125 words) - 16:39, 11 July 2021
- #You will receive 15 points for each correct answer, and 0 points for each problem left unanswered or incorrect. ...rasers. No calculators, smartwatches, or computing devices are allowed. No problems on the test will require the use of a calculator.6 KB (985 words) - 17:39, 11 July 2021
- ==Problem== ...\mod 2</math> must work for <math>xk \le 100</math>. Clearly <math>k=\{2,4,6,8,10 \}</math>, which means the maximum value that <math>xk</math> can take2 KB (269 words) - 20:06, 11 July 2021
- ==Problem== ...id <math>n</math> are <math>4,6,12,</math> leading to an answer of <math>4+6+12=\boxed{22}</math>. ~samrocksnature2 KB (404 words) - 13:00, 12 July 2021
- ==Problem== ...ultiplying these two, we have <math>(ab)^3=12 \cdot 18</math> or <cmath>ab=6 \qquad (3).</cmath> We divide <math>(3)</math> by <math>(1)</math> to get <2 KB (272 words) - 09:07, 12 July 2021
- ==Problem== ...e an odd integer (not necessarily positive) such that <cmath>\dfrac{p^{n+p+2021}}{(p+n)^2}</cmath> is an integer. Find the sum of all distinct possible val2 KB (280 words) - 20:17, 11 July 2021