Last visit was: 15 Jul 2024, 00:41 It is currently 15 Jul 2024, 00:41
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10135
Own Kudos [?]: 17015 [9]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
GMAT Club Legend
Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Posts: 6020
Own Kudos [?]: 13781 [4]
Given Kudos: 125
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3712
Own Kudos [?]: 17313 [1]
Given Kudos: 165
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3712
Own Kudos [?]: 17313 [0]
Given Kudos: 165
Re: If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is [#permalink]
We have posted an article on this concept.

GMATWhiz Representative
Joined: 07 May 2019
Posts: 3401
Own Kudos [?]: 1857 [0]
Given Kudos: 68
Location: India
GMAT 1: 740 Q50 V41
GMAT 2: 760 Q51 V40
Re: If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is [#permalink]
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.
Tutor
Joined: 26 Jun 2014
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
Posts: 450
Own Kudos [?]: 809 [0]
Given Kudos: 8
Re: If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is [#permalink]
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

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10135
Own Kudos [?]: 17015 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: If 100! = 100 x 99 x…x 2 x 1 can be written as 2^a3^b5^c7^d…., what is [#permalink]
1
Bookmarks
=>

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 $$50 + 25 + 12 + 6 + 3 + 1 = 97.$$

Manager
Joined: 04 Jun 2020
Posts: 68
Own Kudos [?]: 159 [0]
Given Kudos: 16
Location: India
Concentration: Strategy, General Management
GPA: 3.4
WE:Engineering (Consulting)
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
Intern
Joined: 09 Feb 2021
Posts: 16
Own Kudos [?]: 0 [0]
Given Kudos: 0
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