N = 200*210*...*290*300 = (10^11)(20*21...29*30).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
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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