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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 [#permalink]
Expert Reply
We have posted an article on this concept.

To know, please go and read : Variations in Factor Manipulation
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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 [#permalink]
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MathRevolution wrote:
[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. \(131\)


Solution


    • \(100! = 100*99*98*……. *2*1\)
    We need to find the power of 2 in 100!

    • Note that 2 is a prime number.

So we can use the below formula to find the power of a prime number in a factorial :
Power of prime number p in \(n! = [\frac{n}{p^1}] + [\frac{n}{p^2}] + [\frac{n}{p^3}] ………………… \) and so on, where n is a positive integer and [ ] denotes greatest integer function.
In our case \( p= 2\) and \(n =100\)
    • The maximum power of 2, that gives a value less than \(100 = 2^6 \)
      o So, for \([\frac{100}{2^x}] = 0\) for all \(x>6\) where x is a positive integer.
      o Hence, to find the value of a, we need not to worry about the powers of 2 which are greater than 6.
Thus, Power of 2 in \(100! = [\frac{100}{2^1}] + [\frac{100}{2^2}] + [\frac{100}{2^3}] + [\frac{100}{2^4}] + [\frac{100}{2^5} ] + [\frac{100}{2^6}] = 50 +25 + 12 + 6 + 3 +1 = 97\)
Thus, the correct answer is Option 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 [#permalink]
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MathRevolution wrote:
[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 [#permalink]
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=>

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 \([\frac{100}{2}] = 50.\)

The number of multiples of \(4\) is \([\frac{100}{4}] = 25.\)

The number of multiples of \(8\) is \([\frac{100}{8}] = 12.\)

The number of multiples of \(16\) is \([\frac{100}{16}] = 6.\)

The number of multiples of \(32\) is \([\frac{100}{32}] = 3.\)

The number of multiples of \(64\) is \([\frac{100}{64}] = 1.\)

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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If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is [#permalink]
100/2 = 50 +
100/4 = 25+
100/8= 12+
100/16= 6+
100/32= 3+
100/64= 1

=97
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If 100! = 100 x 99 xx 2 x 1 can be written as 2^a3^b5^c7^d., what is [#permalink]
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

Answer: B


Claudio Hurtado
Claudio Hurtado, GMAT Quant Prep by https://gmatchile.cl­
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If 100! = 100 x 99 xx 2 x 1 can be written as 2^a3^b5^c7^d., what is [#permalink]
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