VeritasKarishma wrote:
Abhi077 wrote:
If \(\frac{100!}{x}\) is not an integer, which of the following could be x?
A) \(5^{24}\)
B) \(7^{16}\)
C) \(11^9\)
D) \(13^6\)
E) \(17^6\)
The question is based on this concept:
https://www.veritasprep.com/blog/2011/0 ... actorials/Looking at the options, we see that we will need to find the highest power of each prime in 100!
Checking each option would be too much work so let's try to make some intelligent guesses. Looking at options (C), (D) and (E), we see that till 100, we will have only a few multiples of 11, 13 and 17.
11*9 < 100 so we will have 9 11s in 100!
Between 13^6 and 17^6, if we have 6 multiples of 17, we will obviously have 6 multiples of 13 too.
17*6 = 102 so till 100, we have only 5 multiples of 17.
Hence 100! will not be divisible by 17^6.
Answer (E)
Responding to a pm:
Quote:
Can you please elaborate on answer choice b) also, using "17*6 = 102" concept. Thank you.
100! = 1*2*3*4*5*6*7*8* ... *14* ... *21* ... *49*... *98*99*100
Each 7th humber is a multiple of 7 so it gives us a 7:
7, 14, 21, 28,.. 98
These are 14 numbers because 7*14 = 98
But 49 and 98 gives us two 7s each so we get another two 7s.
In all we get 16 7s so 7^16 divides 100! evenly.
Since 5 and 7 are smaller numbers and have numbers with multiple 5s and 7s (25, 50, 49 etc), I chose to ignore those two options to begin with and focus on the larger numbers. Those don't have squares/cubes etc within 100 numbers (11*11 = 121, 13*13 = 169 etc)
The link I have given above tells you how to quickly calculate the highest power of any number in a factorial.
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Karishma
Veritas Prep GMAT Instructor
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