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29 Nov 2021, 03:15
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
57% (02:33) correct
43%
(02:36)
wrong
based on 261
sessions
History
Date
Time
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What is the largest value of x for which \(12^x\) leaves zero remainder when it divides \(\frac{50!}{10!}\)?
A. 9
B. 18
C. 19
D. 22
E. 39
A. 9
B. 18
C. 19
D. 22
E. 39
29 Nov 2021, 04:33
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50!
Number of 3s = \(\frac{50}{3}\) + \(\frac{50}{9}\) + \(\frac{50}{27}\) = 16 + 5+ 1 = 22
Number of 2s = \(\frac{50}{2}\) + \(\frac{50}{4 }\) + \(\frac{50}{8}\) + \(\frac{50}{16}\) + \(\frac{50}{32}\) = 25 +12 + 6 + 3 + 1 = 47. Hence number of 4s = 46/2 = 23
Hence no. of 12s = 22
10!
Number of 3s = \(\frac{10}{3}\) + \(\frac{10}{9}\) = 3 + 1 = 4
Number of 2s = \(\frac{10}{2}\) + \(\frac{10}{4 }\) + \(\frac{10}{8}\) = 5 +2 + 1 = 8. Hence number of 4s = 8/2 = 4
Hence no. of 12s = 4
Hence \(\frac{50!}{10!}\) will contain \(12^{22 - 4} \)= \(12^{18}\)
Number of 3s = \(\frac{50}{3}\) + \(\frac{50}{9}\) + \(\frac{50}{27}\) = 16 + 5+ 1 = 22
Number of 2s = \(\frac{50}{2}\) + \(\frac{50}{4 }\) + \(\frac{50}{8}\) + \(\frac{50}{16}\) + \(\frac{50}{32}\) = 25 +12 + 6 + 3 + 1 = 47. Hence number of 4s = 46/2 = 23
Hence no. of 12s = 22
10!
Number of 3s = \(\frac{10}{3}\) + \(\frac{10}{9}\) = 3 + 1 = 4
Number of 2s = \(\frac{10}{2}\) + \(\frac{10}{4 }\) + \(\frac{10}{8}\) = 5 +2 + 1 = 8. Hence number of 4s = 8/2 = 4
Hence no. of 12s = 4
Hence \(\frac{50!}{10!}\) will contain \(12^{22 - 4} \)= \(12^{18}\)
29 Nov 2021, 12:31
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Bunuel
Prime factorization of 12^x = 2^2*3 (Since repetition of 3 is bigger than 2^2 so we will consider the number of 3's)
Number of 3's in 50! = 50!/3 = 16+5+1 = 22.
Number of 3's in 10! = 10!/3 = 3+1 = 4.
So, 50!/10! = 22-4 = 18
So the largest value of X is 18. B
