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If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is a?

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If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is a?

by Max@Math Revolution » Fri Jan 31, 2020 1:45 am
[GMAT math practice question]

If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is a?

A. 86
B. 97
C. 108
D. 119
E. 131
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Re: If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is a?

by sujoykrdatta » Fri Jan 31, 2020 6:32 am
Max@Math Revolution wrote: ↑
Fri Jan 31, 2020 1:45 am
 [GMAT math practice question]

If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is a?

A. 86
B. 97
C. 108
D. 119
E. 131
This question is basically asking you the highest power of 2 in 100!

To calculate the highest power of a prime p (= 2) in N! (= 100!), we do:
\([\frac{N}{p}]+[\frac{N}{p^2}]+[\frac{N}{p^3}]+...\)

Where [k] refers to the greater integer less than or equal to k

Thus, we have:
\([\frac{100}{2}]+[\frac{100}{2^2}]+[\frac{100}{2^3}]+[\frac{100}{2^4}]+[\frac{100}{2^5}]+[\frac{100}{2^6}]\)
= 50 + 25 + 12 + 6 + 3 + 1 = 97

Answer B
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Re: If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is a?

by Max@Math Revolution » Sun Feb 02, 2020 6:22 pm
=>

We need to count the number of prime factors 2 in the prime factorization of 100!. We can do it by counting:

The number of multiples of 2 is [100/2] = 50.
The number of multiples of 4 is [100/4] = 25.
The number of multiples of 8 is [100/8] = 12.
The number of multiples of 16 is [100/16] = 6.
The number of multiples of 32 is [100/32] = 3.
The number of multiples of 64 is [100/64] = 1.

Thus the number of prime factors 2 is 50 + 25 + 12 + 6 + 3 + 1 = 97.

Therefore, the answer is B.
Answer: B
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