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Greatest possible value of n

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by gmattesttaker2 » Wed Apr 02, 2014 9:24 pm
Hello,

Can you please tell me how to solve this?

If 10^n is a factor of the product of the first 24 positive integers, what is the
greatest possible value of n?

(A) 7
(B) 6
(C) 5
(D) 4
(E) 3

OA: D


Thanks for your help.

Best Regards,
Sri
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by [email protected] » Wed Apr 02, 2014 10:55 pm
Hi Sri,

This question is based on a math concept called Prime Factorization.

We're told that 10^n is a FACTOR of the product of the first 24 positive integers. This means that (2x5)^n is a factor. We need to find all of the "5s" that appear in the first 24 positive integers.

Those numbers include:
5
10 = 2x5
15 = 3x5
20 = 2x2x5

There are four "5s" in the first 24 positive integers, so the greatest "power" of 10^n would be 10^4.

Final Answer: D

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by Abhishek009 » Thu Apr 03, 2014 6:57 am
gmattesttaker2 wrote:Hello,

Can you please tell me how to solve this?

If 10^n is a factor of the product of the first 24 positive integers, what is the
greatest possible value of n?

(A) 7
(B) 6
(C) 5
(D) 4
(E) 3

OA: D


Thanks for your help.

Best Regards,
Sri


Good Problem ,
product of the first 24 positive integers
=> 24 *23 *22 *21 .............. 2 * 1 = 24!
10^n is a factor of the product
Remember 10^n means any number which is a multiple of 10
what is the greatest possible value of n?
To find the greatest possible value of " n " we need to find the number of multiple of 10

Or , Practically speaking we need to count to number of zero's in the product of 24!


So work out -

Number of zero's in 24! is nothing but the number of 5's in it , so calculate the total number of 5's in the product of 24!

This is nothing but -

24/5 => 4 leaving remainder 4


So we have 4 fives in 24!

Hence the maximum value of " n " is 4...
Abhishek
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