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N is the product of the multiples of 10 between 199 and 301

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by faraz_jeddah » Tue Apr 23, 2013 11:10 pm
If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which N / (10)^m is an integer?

a - 10
b - 11
c - 12
d - 14
e - 15

I know there are 10 multiples of 10 between the range.
But then do I find the product of 20 x 21 x 22...30 ?

OA is E
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by Anju@Gurome » Tue Apr 23, 2013 11:50 pm
faraz_jeddah wrote:If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which N/(10)^m is an integer?
Basically, we need to find out the number of zeroes at the end of N.
Because if m is the greatest integer for which N/(10)^m is an integer, there must be m zeroes at the end of N.

Number of multiples of 10 in between 199 and 301 = number of multiples of 10 in between 200 and 300, both inclusive = (300 - 200)/10 + 1 = 100/10 + 1 = 10 + 1 = 11

Otherwise we can list the multiples of 10 in this region also : 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, and 300 ---> 11 multiples of 10

Each of these 11 multiples of 10 will result in one zero at the end of N.
200 and 300 each has one more zero in them --> 2 more zeroes at the end of N
250 has two 5s in it and there are plenty of 2s to make two more zeroes (as 5*2 = 10) ---> 2 more zeroes at the end of N

Hence, total number of zeroes at the end of N = (11 + 2 + 2) = 15

The correct answer is E.
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by Anju@Gurome » Wed Apr 24, 2013 12:06 am
"‹The general approach to solve this type of problem is to find out the number of 5s in the product as 10 is made of one 2 and one 5, and number of 2s will be more than number of 5s in any such product (Hence, number 5s will determine the number of 10s).

"‹"‹Now number of multiples of 5 in the product
= Number of multiples of 5 between 199 and 301 which are multiple of 10 too + Number of multiples of 25 between 199 and 301 which are multiple of 10 too + Number of multiples of 125 between 199 and 301 which are multiple of 10 too
= Number of multiples of LCM(5, 10) between 199 and 301 + Number of multiples of LCM(5, 10) between 199 and 301 + Number of multiples of LCM(125, 10) between 199 and 301
= Number of multiples of 10 between 199 and 301 + Number of multiples of 50 between 199 and 301 + Number of multiples of 250 between 199 and 301
= Number of multiples of 10 between 200 and 300 + Number of multiples of 50 between 200 and 300 + Number of multiples of 250 between 200 and 300
= [(300 - 200)/10 + 1] + [(300 - 200)/50 + 1] + [(300 - 200)/250 + 1]
= [100/10 + 1] + [100/50 + 1] + [100/250 + 1]
= [10 + 1] + [2 + 1] + [0 + 1]
= 11 + 3 + 1
= 15

Note #1 : Number of multiples of 25 and 125 are to be considered as 25 = 5*5 and 125 = 5*5*5. hence, 25 has one extra 5 and 125 has two extra 5s. If the range was much larger we should check for lager powers of 5. Because they will have more extra 5s in them.

Note #2 : The divisions while finding out the number of multiples are integral division only. That means we need to only consider the integral part of the result as number of multiples cannot be fraction. For example, [100/250 + 1] is not equal to usual [0.4 + 1] = 1.4 but integral part of 1.4 = 1
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by GMATGuruNY » Wed Apr 24, 2013 4:09 am
faraz_jeddah wrote:If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which N / (10)^m is an integer?

a - 10
b - 11
c - 12
d - 14
e - 15

I know there are 10 multiples of 10 between the range.
But then do I find the product of 20 x 21 x 22...30 ?

OA is E
N = 200*210*...*290*300 = (10^11)(20*21...29*30).
10^m = the number of 10's that can divide into N.

The number of 10's that can divide into 10^11 = 11.
We need to count how many 10's can divide into the red portion.
Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of the red portion will allow an additional 10 to divide into N.
The prime-factorization of the red portion will be composed of FAR MORE 2'S than 5's.
Thus, to determine HOW MANY 10'S can divide into the red portion, we must count THE NUMBER OF 5'S contained within the prime-factorization of the red portion.

Every multiple of 5 provides at least one 5:
20*25*30 --> three 5's.
Since every multiple of 5² is composed of TWO 5's, every multiple of 5² provides a second 5:
25 --> one more 5.
Thus, the total number of 5's contained within the prime-factorization of the red portion = 3+1 = 4.

Thus, the greatest number of 10's than can divide into N = 11+4 = 15.

The correct answer is E.

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