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# If 100! = 100 x 99 x...x 2 x 1 can be written as (2^a)(3^b)(5^c)(7^d)., what is a?

Monday 15 April 2024,

[GMAT math practice question] If \(100! = 100 x 99 xx 2 x 1\) can be written as \((2^a)(3^b)(5^c)(7^d)\)., what is \(a\)? A. \(86\) B. \(97\) C. \(108\) D. \(119 \) E. \(131\)

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If 100! = 100 x 99 x...x 2 x 1 can be written as (2^a)(3^b)(5^c)(7^d)., what is a?

Explanation by Claudio Hurtado, GMAT Quant Prep by gmatchile.cl:

Since a is the exponent (power) of the base 2, we need to determine how many times the base 2 appears in the factorization of 100!.

There are various methods to do this.

We will first count all the even numbers (multiples of 2) up to 25, considering that

100=4×25 and 2×25=50, 4×24=96, and 2×24=48, and so on.

Each number up to 25 contributes with 2^3 to form numbers up to 100, in addition to the powers of 2 that are formed up to 25.

That is, 25×3=75. We must add the 2s that appear in the factorization up to 25. Let’s see: 25 does not contribute with 2, meaning all odd numbers up to 25 do not contribute 2. Therefore, we only need to consider the even numbers up to 25.

Thus:

24=2×2×2×3=…×2^3

22=2×11=…×2^1

20=2×2×5=…×2^2

18=2×9=…×2^1

16=2×2×2×2=2^4

14=2×7=…×2^1

12=2×2×3=…×2^2

10=2×5=…×2^1

8=2×2×2=2^3

6=2×3=…×2^1

4=2×2=2^2

2=2^1

So we have:

75+3+1+2+1+4+1+2+1+3+1+2+1=97