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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
- == Problem==1 KB (200 words) - 00:35, 20 January 2024
- ==Problem==815 bytes (114 words) - 11:32, 5 July 2013
- == Problem == ...tion with <math>\overline{CN}</math> as the point <math>Q</math>. From the problem, the area of <math>\triangle QMO</math> is <math>n</math>, but by vertical2 KB (415 words) - 19:21, 3 April 2023
- ==Problem== Let <math>P</math> <math>=</math> original population. Translating the word problem into a system of equations, we got:4 KB (570 words) - 15:52, 28 June 2023
- ==Problem== ...>, <math>600</math> feet is insignificant and can be safely covered by the problem's use of the word "approximately." Therefore, the answer is <math>\boxed{\t1 KB (195 words) - 00:01, 9 January 2017
- == Problem ==1,018 bytes (168 words) - 00:40, 16 August 2023
- ==Problem== Plugging in the numbers given in the problem, we get848 bytes (121 words) - 17:17, 9 March 2020
- == Problem == ...is an integer divisible by <math>8</math>. The number of solutions to this problem is:1 KB (161 words) - 00:19, 22 December 2015
- == Problem ==873 bytes (137 words) - 22:23, 13 March 2015
- == Problem ==1 KB (197 words) - 23:48, 5 April 2024
- ==Problem==2 KB (303 words) - 21:27, 28 December 2023
- == Problem ==3 KB (545 words) - 03:29, 17 October 2022
- == Problem ==2 KB (292 words) - 01:08, 20 June 2018
- ==Problem==1 KB (176 words) - 16:08, 15 March 2021
- ==Problem 38==433 bytes (66 words) - 00:42, 4 February 2020
- == Problem ==940 bytes (143 words) - 15:48, 21 July 2024
- ==Problem== ...r <math>N</math> with its digits reversed is <math>10b+a</math>. Since the problem asks for a positive number as the difference of these two numbers, than <ma1 KB (220 words) - 07:35, 27 July 2024
- == Problem == |before=[[1963 TMTA High School Algebra I Contest Problem 37| Problem 37]]1 KB (183 words) - 11:39, 2 February 2021
- == Problem 38==1 KB (176 words) - 22:35, 2 January 2024
Page text matches
- == 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=\fra2 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 fr8 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 works2 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} \ri5 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 number8 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 distinct8 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 = 2X3 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 \cdo6 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 th2 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