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- ==Problem== #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]1 KB (178 words) - 09:41, 12 July 2021
- ==Problem== ...es are equal. Thus, each side of the octagon must equal to <math>\frac{90}{3} = 30</math>. Since one side of the square shares a side with the octagon,1 KB (170 words) - 16:23, 11 July 2021
- ==Problem== ...math>5x < 100 < 6x</math>. We solve the inequality to find <math>\frac{50}{3} \leq x \leq 20</math>, but since <math>x</math> must be an integer we have2 KB (269 words) - 20:06, 11 July 2021
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- ==Problem== If <math>a : b : c : d=1 : 2 : 3 : 4</math> and <math>a</math>, <math>b</math>, <math>c</math>, and <math>d<2 KB (296 words) - 00:17, 12 July 2021
- ==Problem== ...s, or <math>$2.75</math> into <math>$2.78</math>. That only leaves <math>3</math> cents. We cannot put any nickels nor dimes, therefore we require thr1 KB (195 words) - 09:09, 12 July 2021
- ==Problem== ==Solution 3==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 == 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== Case 2: <math>x+y=3, xy=4</math>.2 KB (300 words) - 16:15, 12 July 2021
- == Problem == Find the sum of all positive multiples of <math>3</math> that are factors of <math>27.</math>937 bytes (127 words) - 16:22, 11 July 2021
- ==Problem== ...ctangle can be written as a fully simplified fraction <math>\frac{a+b\sqrt{3}}{c}</math> such that <math>gcd(a,b,c)=1.</math> Find <math>a+b+c</math>.2 KB (333 words) - 16:24, 11 July 2021
- ==Problem== ...h> For example, <math>f(1)=475</math>, <math>f(2)=4775</math>, and <math>f(3)=47775.</math> Find the last three digits of <cmath>\frac{f(1)+f(2)+ \cdots2 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== ...of positive integers that satisfies <cmath>\frac{1}{2}<\frac{2}{x_1}<\frac{3}{x_2}< \cdots < \frac{11}{x_{10}},</cmath> find <math>x_1+x_2+\cdots+x_{10}1 KB (178 words) - 16:24, 11 July 2021
- ==Problem== From the problem, we know that1 KB (205 words) - 16:23, 11 July 2021
- ==Problem== ...satisfy the equation <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== ...triangles about the right side, <math>3</math> triangles containing <math>3</math> sides, <math>2</math> triangles in the middle, and <math>1</math> tr718 bytes (98 words) - 16:23, 11 July 2021
- ==Problem== ...es are equal. Thus, each side of the octagon must equal to <math>\frac{90}{3} = 30</math>. Since one side of the square shares a side with the octagon,1 KB (170 words) - 16:23, 11 July 2021
- == Problem == ...98</math>, and <math>96</math>. Thus, our answer is <math>\frac{100+98+96}{3} = \boxed{98}</math>.980 bytes (133 words) - 12:11, 13 July 2021
- ==Problem== ...rtner (2 houses 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}.</cmat1 KB (197 words) - 09:01, 12 July 2021
- ==Problem== Consider the equilateral triangle that the circles make. There are <math>3</math> ways to pick the circles so that the row they make is parallel to ea1 KB (207 words) - 09:53, 16 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