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  • == Problem 38== In order to see the rest of the problem set, click [[1955 AHSME Problems|here]]
    1 KB (200 words) - 22:35, 28 August 2020
  • * <math>38! = 523022617466601111760007224100074291200000000</math> ([[2007 iTest Problems/Problem 6|Source]])
    10 KB (809 words) - 15:40, 17 March 2024
  • == Problem 1 == [[2004 AMC 12A Problems/Problem 1|Solution]]
    13 KB (1,953 words) - 23:31, 25 January 2023
  • == Problem == ...\frac{19}{6}</math>, so the coordinate of this point is <math>\left(-\frac{38}{5},0\right)</math>. Hence the equation of <math>L_2</math> is <math>y=\fra
    2 KB (295 words) - 01:24, 30 October 2024
  • == Problem == ...to be 4 odds, 4 evens, or 2 odds and evens. The 4 odd numbers do not work (38) nor the 4 even numbers (26).
    3 KB (584 words) - 00:17, 17 July 2024
  • == Problem == With these particular numbers, when converting back to decimal note that the 38 zeroes all cancel out. Let the first 1 when reading the number in binary fr
    8 KB (1,283 words) - 18:19, 8 May 2024
  • == Problem == ...So the probability that they both pick two red candies is <math>\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323}</math>. The same calculation works
    2 KB (330 words) - 12:42, 1 January 2015
  • ==Problem== ...\cdot 2 + 1 = 38</math> ones that form <math>P</math>, so <math>P = \frac{38}{380} = \frac{1}{10}</math>. Therefore the answer is <math>\frac{9}{20} \ri
    5 KB (830 words) - 21:15, 28 December 2023
  • == Problem 1 == [[2000 AIME I Problems/Problem 1|Solution]]
    7 KB (1,204 words) - 02:40, 4 January 2023
  • == Problem == ...math> into two addends, we see that no pair of odd composites add to <math>38</math>. Therefore, <math>\boxed{038}</math> is the largest possible number
    8 KB (1,365 words) - 14:38, 10 December 2024
  • == Problem == ...e can partition as <math>1+2+3</math>, from which we find that <math>\frac 38</math> works.
    12 KB (1,859 words) - 17:16, 28 March 2022
  • == Problem == Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct
    8 KB (1,269 words) - 09:55, 26 June 2024
  • == Problem == In addition, we see that the [[perimeter]] of the rectangle is <math>2AB + 38 = XA + AD + DW + WZ + ZC + CB + BY + YX = 4XY</math>, so <math>AB + 19 = 2X
    3 KB (530 words) - 06:46, 1 June 2018
  • == Problem == ...uence. It is easy to memorize that <cmath>17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.</cmath> Also, <cmath>42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdo
    6 KB (893 words) - 07:15, 2 February 2023
  • == Problem == f(14,52) &= \frac{52}{38} \cdot f(14,38) \
    4 KB (538 words) - 12:24, 12 October 2021
  • == Problem == Consequently, <math>S = - 14</math> and <math>P = - 38</math>. Finally:
    4 KB (644 words) - 15:24, 28 May 2023
  • == Problem == ...lues, with <math>7</math> being the value of the remaining <math>73 - 35 = 38</math> numbers. The 39th number is <math>\lfloor r+\frac{19 + 39 - 1}{100}\
    3 KB (447 words) - 16:02, 24 November 2023
  • == Problem == We know that <math>n^2 = 2^{62}3^{38}</math> must have <math>(62+1)\times (38+1)</math> [[factor]]s by its [[prime factorization]]. If we group all of th
    2 KB (407 words) - 07:14, 4 November 2022
  • == Problem == ...{2} + \frac{\frac{12}{\sqrt{57}}}{2 \cdot \frac{\sqrt{57}}{2}} = \frac{27}{38}</math>, and <math>m+n = \boxed{065}</math>.
    3 KB (521 words) - 00:18, 25 February 2016
  • == Problem == We know that <math>38 \times 9=342</math> and <math>38 \times 10 \equiv 20 \pmod{360}</math> (by simple arithmetic).
    5 KB (757 words) - 20:59, 23 December 2024

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