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- ==Problem== ...ents by <math>s</math>, and number of classes by <math>c</math>. Thus, our problem implies that1 KB (196 words) - 16:24, 11 July 2021
- ==Problem== ...incorrect or skipped question. Find the sum of all the possible numbers of problems that the test could have had.1 KB (154 words) - 16:14, 11 July 2021
- ==Problem== #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]1 KB (169 words) - 18:48, 23 September 2024
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- ==Problem== <cmath>16000\cdots \times 25000\cdots = 16 \times 25 \times 10^{198} = 400 * 10^{198}</cmath>2 KB (209 words) - 16:25, 11 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== ...</math> Using our inductive assumption, we obtain <cmath>\frac{f(n+1)}{25}=10 \cdot (19\underbrace{111 \cdots 1}_{n-1 \text{ ones}})+1=19 \cdot \underbra2 KB (304 words) - 00:19, 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== ...{3}{x_2}< \cdots < \frac{11}{x_{10}},</cmath> find <math>x_1+x_2+\cdots+x_{10}</math>.1 KB (178 words) - 16:24, 11 July 2021
- ==Problem== ...on that this operation is in base <math>10</math>, there exists only <math>10</math> digits to be used, specifically only <math>5</math> for the first di1 KB (194 words) - 09:43, 12 July 2021
- ==Problem== ...s are already selected) The answer is <cmath>\frac{2^2 \cdot 3 \cdot 5}{2^{10}} \implies \frac{15}{256} \implies 256+15=\boxed{271}.</cmath> ~Geometry2851 KB (197 words) - 09:01, 12 July 2021
- ==Problem== ...y numbers with <math>0</math>'s in their digits are the multiples of <math>10</math>.1 KB (147 words) - 09:44, 12 July 2021
- ==Problem== Find the last two digits of <math>10^{10}-5^{10}.</math>984 bytes (129 words) - 19:54, 7 September 2021
- ==Problem== ...area of triangle of <math>ADC</math> is <math>\frac{20 \cdot \sqrt{26^2 - 10^2}}{2} = 240</math>, and the area of triangle <math>ABC</math> is <math>\fr1 KB (148 words) - 09:39, 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== ...ans the maximum value that <math>xk</math> can take on is <math>90=9 \cdot 10</math>, and the minimum value it can take on is <math>2=2 \cdot 1</math>. S2 KB (269 words) - 20:06, 11 July 2021
- ==Problem== .... If <math>\angle ADC=2\angle ABC</math>, <math>AD=13</math>, and <math>BC=10</math>, find <math>AC.</math>1 KB (154 words) - 17:10, 11 July 2021
- ==Problem== ...h> or <math>9</math> are part of our sequence, so for every cycle of <math>10</math> perfect squares, exactly <math>4</math> are included. This means <ma3 KB (435 words) - 20:15, 11 July 2021
- ==Problem== pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */7 KB (744 words) - 20:18, 11 July 2021