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- ==Problem== #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]1 KB (178 words) - 16:24, 11 July 2021
- ==Problem== #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]797 bytes (121 words) - 16:14, 11 July 2021
- ==Problem== #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]1 KB (154 words) - 17:10, 11 July 2021
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- ==Problem== ...>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>\{21,42,2 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
- ==Problem== Therefore, the answer is <math>1 + 3 + 0 + 5 = 9</math>.2 KB (300 words) - 16:15, 12 July 2021
- == Problem == .../math>, <math>9</math>, and <math>27</math>. Thus, our answer is <math>3 + 9 + 27 = \boxed{39}</math>.937 bytes (127 words) - 16:22, 11 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== ...n <cmath>\begin{array}{cccc}& A & B & C \ \times & & &3 \ \hline & 7 & 9 & C\end{array},</cmath> find <math>3A+2B+C</math>.2 KB (332 words) - 09:42, 18 July 2021
- ==Problem== ...lly only <math>5</math> for the first digit. Only <math>8</math> and <math>9</math> may be used, as there isn't other pair of digits which sum to <math>1 KB (194 words) - 09:43, 12 July 2021
- ==Problem== ...lel to each side of the equilateral triangle. Thus, there are <math>\boxed{9}</math> ways total. [diagram needed]1 KB (207 words) - 09:53, 16 July 2021
- ==Problem== ...re <math>100-11+1=90</math> two-digit numbers total, so there are <math>90-9=\boxed{81}</math> numbers.1 KB (147 words) - 09:44, 12 July 2021
- ==Problem== ...y numbers are in the finite sequence of consecutive perfect squares <cmath>9, 16, 25, \ldots , 2500?</cmath>1 KB (168 words) - 18:55, 7 September 2021
- ==Problem== ...t of both sides gives <math>y=2^9,</math> thus we have <math>\sqrt[4]{x}=2^9,</math> which makes <math>x=2^{36}.</math> Answer is <math>\boxed{36}.</mat2 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== ...the integers <math>x</math> must be <math>9</math>. The sum is <math>18+20+9</math> or <math>\boxed{47}</math>2 KB (269 words) - 20:06, 11 July 2021
- ==Problem== ...ose all the squares are NOT fully covered by shaded squares. We have <math>9</math> total cases that include this. Now, the center square already covers2 KB (245 words) - 14:09, 6 August 2021
- ==Problem== For some <math>n</math>, the arithmetic progression <cmath>4,9,14,\ldots,n</cmath> has exactly <math>36</math> perfect squares. Find the m3 KB (435 words) - 20:15, 11 July 2021
- ==Problem== ...2 + 4^2 = 9 + 16 = 25</math>, while <math>CE^2 = CG^2 + GE^2 = 3^2 + 7^2 = 9 + 49 = 58</math>. Thus, <math>CF^2 + CE^2 = 25 + 58 = \boxed{83}.</math> ~B3 KB (425 words) - 20:45, 22 December 2021