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by guerrero » Fri Apr 12, 2013 1:53 pm
If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which N/10^m is an integer?

A. 3
B. 6
C. 7
D. 8
E. 10


I am unable to follow what the question is testing about ? Appreciate if someone can simplify the statement .

OA C
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by Zarrolou » Fri Apr 12, 2013 2:12 pm
Basically the question asks how many 5*2 there are in N, so that N is divisible by (5*2)^m
The problem now comes down to how many 5s there are in N, because clearly the number of 2s will be greater so we need to find the 5s to make the combination 5*2.
15,30,45,60,75,90 have 5. How many 5s do they contain?
All they contain one 5 (ie 30=2*3*5) except 75 that contains two 5s (75=5*5*3); so tot number of 5s is 7.
C
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by Anju@Gurome » Fri Apr 12, 2013 9:32 pm
guerrero wrote:If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which N/10^m is an integer?
For N/10^m to be an integer, 10^m must divide N.

Now, 10^m is a power 10.
So, if N is divisible by 10^m, N must contain m number of 10s, i.e. m number of 2s and 5s.

Now, as N = product of all multiples of 3 between 1 and 100, there will be more 2s than 5s in N.
So, it is easier for us to calculate the number of 5s in N.

Number of 5s in N = Number of 5s in 15, 30, 45, 60, 75, and 90 = 1 + 1 + 1 + 1 + 2 + 1 = 7

So, N at the most contain seven 5s.
--> N at the most will be divisible by 10^7.
--> Greatest value of m is 7.

The correct answer is C.
Anju Agarwal
Quant Expert, Gurome

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