If the number 200! is written in the form p × 10^q, where p and q are

Question

If the number 200! is written in the form p × 10^q, where p and q are integers, what is the maximum possible value of q?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64

GMATinsight · Accepted Answer

We have to find the maximum power of 10 available in 200!

10 is made up of 2 x 5

i.e. we have to find the powers of 2 and 5 to find the powers of 10 in 200!

but powers of 2 in 200! will be greater than power of 5 because there are more numbers multiple of 2 (100 Numbers) than multiple of 5 (40 Numbers)

hence the power of 5 will be the power of 10 in 200!

to calculate the power of 5 in 200! =   +   +

here   represents the greatest integer less than or equal to x

hence, Power of 5 in 200! = 40 + 8 + 1 = 49

Answer: Option  C

tjerkrintjema · Answer

I thought this was a really hard question, because at first I was thinking how can you write 200! in this way.
But then i thought that this is about finding the maximum number of time you can divide by 10 (basically how many zeroes) 200! has.
So when you think about 10 one important prime factor is 5 and the other is 2.
Of course 200 has many more 2s in it, than 5s, so this will be our limiting constraint.
200/5 = 40
200/25 = 8
200/125 = 1
= 49

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION: To maximize the value of q, we need to constrain p to NOT be a multiple of 10. In other words, q should be the count of every factor of 10 in 200!, and p would simply be the product of the remaining factors of 200!. To count factors of 10 in 200!, we could start by counting the multiples of 10 between 1 and 200, inclusive. But this method would undercount the number of 10's that are factors of 200!. For example: 200! has 2 and 5 as factors, which multiply to 10, and thus another factor of 10. 200! has 6 and 15 as factors, which multiply to 90, and thus another factor of 10. 200! has 8 and 125 as factors, which multiply to 1000, and thus three more factors of 10. Therefore, this problem is really about counting the number of 2 × 5 factor pairs that can be made from the factors of 200. Let's count the number of 5s found among the factors of 200!: 2015-06-08_1557.png We have counted not only the multiples of 5, but also the multiples of 25 and 125, as these higher multiples contribute more than one factor of 5 to the total. There are eight multiples of 25 in our range (namely, 25, 50, 75, 100, 125, 150, 175, and 200). Each of these are also multiples of 5, so we have already counted one factor of 5 for them, and thus each of the eight multiples of 25 contributes one additional factor of 5. Finally, 125 contributes a total of three 5s to our count — but we already counted one factor of 5 when we counted multiples of 5, and another when we counted multiples of 25, leaving only one additional factor of 5 that we need to count for 125. Thus, the prime factorization for 200! includes 40 + 8 + 1 = 49 factors of 5. It is fairly easy to see that there are many more factors of 2 in 200!, as there are 100 even numbers between 1 and 200, not to mention the additional 2's coming from the multiples of 4, 8, 16, 32, etc. that contribute extra factors of 2. Thus, we can only make 49 of the 2 × 5 factor pairs, and the maximum value of q is 49. The correct answer is C.

Bunuel · Answer

For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory . Theory on Trailing Zeros: everything-about-factorials-on-the-gmat-85592.html

ujjwal80 · Answer

This is a trailing zero concept question: https://gmatclub.com/forum/everything-a ... 85592.html # of trailing zeros 200! has is 49. Therefore, the answer is C.

bumpbot · Answer

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If the number 200! is written in the form p × 10^q, where p and q are

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If the number 200! is written in the form p × 10^q, where p and q are

Prep Toolkit

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