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  • ...ath>c</math>, and <math>d</math> are divisors of <math>252</math>, what is the maximum value of <math>a</math>? ...mid 252</math>. Thus, we must have <math>12a \mid 252</math>. This implies the maximum value of <math>a</math> is <math>252/12 = \boxed{21}</math>, which
    2 KB (296 words) - 00:17, 12 July 2021
  • ...($0.25), dimes ($0.10), nickels ($0.05), and pennies ($0.01). What is the least number (quantity, not type) of coins Brady can use to pay off $<math ...the most value, specifically the quarters, for taking away a big chunk of the problem. We are able to fit <math>11</math> quarters, or <math>$2.75</math
    1 KB (195 words) - 09:09, 12 July 2021
  • ...and all angles marked with one black curve are equal, find the measure of the angle with a question mark. ...l the angles in the two triangles are all equal. We can already infer that the black angles are all <math>60</math> degrees, since they are equilateral tr
    1 KB (178 words) - 09:41, 12 July 2021
  • ...d at square E4 (row 4, column E). How many squares can be attacked by both the bishop and rook if they each take one move? ...uding squares A7, F8, G1, D6, and so on. The rook can attack any square in the same row or column as itself, including squares E5, E8, E4, C4, and so on.)
    1,017 bytes (175 words) - 16:12, 11 July 2021
  • What is the leftmost digit of the product <cmath>\underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \ We conclude that the leftmost digit must be <math>\boxed{4}</math>.
    2 KB (209 words) - 16:25, 11 July 2021
  • '''2021 JMPSC Accuracy (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
  • <b>2021 JMPSC Sprint 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
  • ...he left side gives <math>\frac{18}{3}+\frac{18}{6}+\frac{18}{9}</math>, so the answer is <math>6 + 3 + 2 = \boxed{11}</math>. ...rac{3}{18}+\frac{2}{18}=\frac{11}{18}</math> and <math>3+6+9=18</math>, so the answer is <math>\frac{11}{18}\cdot18=\boxed{11}</math>.
    4 KB (634 words) - 09:53, 12 July 2021
  • ...th> be nonnegative integers such that <math>(x+y)^2+(xy)^2=25.</math> Find the sum of all possible values of <math>x.</math> The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 + 5</math>, which are not real.
    2 KB (300 words) - 16:15, 12 July 2021
  • Find the sum of all positive multiples of <math>3</math> that are factors of <math>2 ...fact that <math>3 = 3^1</math> and <math>27 = 3^3</math> to conclude that the only multiples of <math>3</math> that are factors of <math>27</math> are <m
    937 bytes (127 words) - 16:22, 11 July 2021
  • ...with height <math>1</math> is also inscribed in the triangle. The base of the second rectangle can be written as a fully simplified fraction <math>\frac{ ...{\sqrt{3}}{3}.</cmath> Since we are looking for the base <math>d</math> of the horizontal rectangle and we are given <cmath>PA=1,</cmath> we have <cmath>d
    2 KB (333 words) - 16:24, 11 July 2021
  • ...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)+ \cdots + f(100)}{25}.</cmath> ...h>20+28(1(10)+99+10(98)+100(97) \cdots)+4(444444 \cdots )</cmath> Evaluate the last <math>3</math> digits to get <cmath>20+28(10+99+980+700)+4(444)=\boxed
    2 KB (304 words) - 00:19, 12 July 2021
  • #You have 60 minutes to complete the test. ...es, 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
  • ...</math> as <math>4,5,6,...,13</math> as an example. The first few terms of the inequality would then be: ...we have the same case, <math>x_1=3</math>, <math>x_2=4</math>, and so on. The answer is <math>\frac{12(13)}{2}-3=\boxed{75}</math>
    1 KB (178 words) - 16:24, 11 July 2021
  • ...by <math>1</math>, then each class would have <math>20</math> students. If the number of classes was to decrease by <math>1</math>, then each class would Let the number of total students by <math>s</math>, and number of classes by <math>
    1 KB (196 words) - 16:24, 11 July 2021
  • If <math>\frac{x+2}{6}</math> is its own reciprocal, find the product of all possible values of <math>x.</math> From the problem, we know that
    1 KB (205 words) - 16:23, 11 July 2021
  • ...-2) \cdots 2 \cdot 1</math> for all positive integers <math>n</math>. Find the value of <math>x</math> that satisfies <cmath>\frac{5!x}{2022!}=\frac{20}{2 We can multiply both sides by <math>2022!</math> to get rid of the fractions
    804 bytes (106 words) - 09:04, 12 July 2021
  • ...math>\angle BCD</math>. If <math>AB=15</math> and <math>BC=13</math>, find the perimeter of <math>ABCD</math>. Therefore, the perimeter is <cmath>15+15+13+13=\boxed{56}</cmath>
    1 KB (164 words) - 16:53, 11 July 2021
  • ...</math>, and <math>C</math> each represent a single digit and they satisfy the equation <cmath>\begin{array}{cccc}& A & B & C \ \times & & &3 \ \hline ...th>AB5=\frac{795}{3}=265</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
  • ...ed by segments in the figure and contain the red triangle? (Do not include the red triangle in your total.) ...in the middle, and <math>1</math> triangle containing the right-two sides. The answer is <math>\boxed{8}</math>
    718 bytes (98 words) - 16:23, 11 July 2021

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