What is the greatest positive integer n such that 5^n divides 10! – (2

Question

What is the greatest positive integer n such that  5^n  divides  10! – (2)(5!)^2  ?

A. 2
B. 3
C. 4
D. 5
E. 6

PS14051.02

nick1816 · Accepted Answer

10!- 2(5!)^2  =  5!

=  5!

= 2*5!*5!(126-1)

= 2*5!*5!*5^3

=  2*4!*5*4!*5*5^3

n= 1+1+3=5

BrentGMATPrepNow · Answer

The question is basically asking us to determine how many 5's we can factor out of  10! – (2)(5!)^2

This calls for some  prime factorization! 
--------------ASIDE---------------------
Here's a quick technique for finding the prime factorization of a factorial.

For this example, we'll find the prime factorization of 10!

First, I know that the prime numbers "hiding" in 10! will range from 2 to 7 (i.e., all primes less than or equal to 10), which means we can write:  10! = (2^?)(3^?)(5^?)(7^?) 
From here, I'll focus on one prime number at a time, starting with  2 . 
In 10!, we have: 
 1   2  hiding in 2
 2   2 's hiding in 4 (since 4 = 2x2)
 1   2  hiding in 6 (since 6 = 2x3)
 3   2 's hiding in 8 (since 8 = 2x2x2)
 1   2  hiding in 10 (since 10 = 2x5)
So, there's a total of  8   2 's hiding in 10!, which means we can now write:  10! = (2^8)(3^?)(5^?)(7^?)

Now let's focus on how many  3 's are hiding in 10!. 
We have: 
 1   3  hiding in 3
 1   3  hiding in 6 (since 6 = 2x3)
 2   3 's hiding in 9 (since 9 = 3x3)
So, there's a total of  4   3 's hiding in 10!, which means we can write:  10! = (2^8)(3^4)(5^?)(7^?)

Onto the  5 's: 
 1   5  hiding in 5
 1   5  hiding in 10 
So, there's a total of  2   5 's hiding in 10!, which means we can write:  10! = (2^8)(3^4)(5^2)(7^?)

Finally, we can see that there's exactly  1   7  hiding in 10!, which means we can write:  10! = (2^8)(3^4)(5^2)(7^1) 
-------BACK TO THE QUESTION----------------------------

Following the same procedure for  5! , we get:  10! – (2)(5!)^2     = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 - 2(2^3 \cdot 3 \cdot 5)^2

Simplify the right side:   = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 - 2(2^6 \cdot 3^2 \cdot 5^2)

Simplify more:   = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 - 2^7 \cdot 3^2 \cdot 5^2

Factor:   = 2^7 \cdot 3^2 \cdot 5^2(2 \cdot 3^2 \cdot 7 - 1)

Evaluate the part in parentheses:   = 2^7 \cdot 3^2 \cdot 5^2(125)

Rewrite  125  as a power of  5  to get:   = 2^7 \cdot 3^2 \cdot 5^2(5^3)

Simplify one last time:   = 2^7 \cdot 3^2 \cdot 5^5

So, in total, we can factor  five   5 's out of  10! – (2)(5!)^2

Answer: D

Cheers,
Brent

lacktutor · Answer

10! = 5!* 6*7*8*9*10 = 5!*(2*3)*7*(2*4)*9(2*5)= (5!)^{2}* 252

-->  10! – (2)(5!)^2= (5!)^{2}* 252 - (2)(5!)^2= (5!)^2 (252-2)= 250*(5!)^2= 2*5^{3}*(5!)^{2} 
 5^{n} =5^{5} 
-->  n =5

Answer (D).

chetan2u · Answer

10! – (2)(5!)^2=5!(10*9*8*7*6-2*5!)

5!(2*5*3*3*2*2*2*7*2*3-2*2*3*2*2*5)=5!(2^5*3^3*5*7-2^4*3*5)=

=5!*2^4*3*5(2*3^2*7-1)=5!*2^4*3*5(126-1)=

5*4!*2^4*3*5*125=2^4*3*5^{1+1+3}*4!

So we have five 5s in the terms, and the maximum value of n is 5

D

GMATinsight · Answer

10! – (2)(5!)^2 = 5!*(6*7*8*9*10 - 2*120) = 5!*240*(7*2*9-1) = 5^2*24*48*(125) = 5^5*24*48

i.e. if  5^n  is divisor of  10!-2*5!^2  then,   n_{max} = 5

Answer: Option D

Louis14 · Answer

10! - 2(5!)(5!) 10*9*8*7*6*5! - 2(5!)(5!) 5! 10 *9* 8 *7*6 - 2 ( 5 * 4 *3* 2 )] 5! 5! 5! Now, 5! has one 5; 80 has one 5; 375 has three 5s. So, the max number of 5s are 5 i.e. n = 5 Kudos = Thank you.

Ankyta333 · Answer

Alternate Method:

If you know the values up to 10!, you can solve this question using the actual values as follows:
10! = 3628800 & 5!= 120.
Now the expression is 10!–(2)(5!)^2
We need 2 * Square of (5!) = 2* Square of 120 = 2* 14400= 28800

So, expression becomes 3628800-28800= 36,00,000 Since there are five zeroes, the maximum power of 5 that can divide it will be five.
And hence n=5.

Sa800 · Answer

what is this testing and where can i find topics on what this is testing?  Bunuel

Bunuel · Answer

Check the following topics from our  Special Questions Directory :

12.  Trailing Zeros 
13.  Power of a number in a factorial

Hope it helps.

woohoo921 · Answer

chetan2u 
Thank you for your helpful reply. I am not sure where I went wrong with my method.

I wrote out the 10 factorial as follows:
10*9*8*7*6*5*4*3*2 --> you get two 5s from 10 factorial

For (2)(5!)^2
I did 2*  --> you get two 5s from this.

Where am I missing a 5? Thanks for your help!

chetan2u · Answer

When you subtract the remaining, you will get more 5s.

Simpler example. 
 150-25=2*3*5^2-5^2 
So you would expect  5^2  but 
 150-25=2*3*5^2-5^2=5^2(2*3-1)=5^2*5

KarishmaB · Answer

Here is a discussion on how to solve highest power in factorials: https://anaprep.com/number-properties-h ... actorials/ The things to keep in mind with factorial terms that are added/subtracted - take common and fully solve what is leftover 10! – (2)(5!)^2 = 2 * 5! * (10*9*8*7*3 - 5!) = 2 * 5! * (10*9*8*7*3 - 120) = 2 * 5! *120 (9*2*7 - 1) = 2 * 5! *120 * 125 So we get one 5 from 5!, one 5 from 120 and 3 5s from 125 to give us total 5 5s. Answer (D)

patronumbeagle · Answer

Hi  KarishmaB , please, can you explain me what am i doing wrong?

10!-2(5!)(5!)
= 5!x 
= 5! x  
= 5!x10x8x6  
= 5!x10x8x6

5! have one 5 and 10 have one 5, so max n is 2

Bunuel · Answer

The error is in line 3:

2(5 * 4 * 3 * 2) does not equal 10 * 8 * 6 * 4. It's not a sum in the parentheses, so you do not apply multiplication by 2 to each term. You can get, say, 10 * 4 * 3 * 2.

ablatt4 · Answer

The most efficient way to solve this question is to invest the time to break it down before rushing into a bad strategy.

The questions is effectively asking us "how many times does 5 factor out of this quantity" meaning whatever amount of 5s we can factor out is the answer, much more simple already.

Next, we need to identify that rushing into solving these massive numbers is certainly doable, but not in a timely manner.

Let's see what can be factored.

Term 1 is 10! or 10*9*8*7*6*5*4*3*2 which can be reduced into 5^2*2^8*7*3^4

Term 2 is 2(5!) or 2^7*3*2*5^2

From each we can factor out 5^2*2^7*3^2 and have (2*7*3^2 - 1) left

14*9=126-1=125= 5^3

Leaving us with 5^2*2^7*3^2*5^3

5^2*5^3= 5^5 OR we can factor 5 5s from this equation therefore D is the answer

Quick recap.

we are factoring as many 5s out of a quantity as we can.

Quantity is too big to solve so we simplified into prime factors and factored out of each term in the quantity.

we were then able to solve for the amount of 5s too factor out.

What is the greatest positive integer n such that 5^n divides 10! – (2

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What is the greatest positive integer n such that 5^n divides 10! – (2

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