How many factors does N = 2^3*3^2*5^4 have that end with zero?

Question

How many positive factors does  N = 2^3*3^2*5^4  have that end with zero?

A. 24 
B. 36 
C. 48
D. 50 
E. 60

GMATinsight · Accepted Answer

N = 2^3*3^2*5^4

FOr having zero at the end of a number, the result must be a multiple of 10   therefore

N = (10)*2^2*3^2*5^3

CONCEPT:  Number of Factors
  For a Number  N=a^p*b^q*c^r*...  where a, b, c etc are distinct prime factors of n and p, q r are their respective exponents
The number of factors of  N = (p+1)*(q+1)*(r+1)*...  and so on

So keeping 10 aside, the total factors of  \frac{N}{10} = 2^2*3^2*5^3  will be (2+1)*(2+1)*(3+1) = 36

i.e. N/10 has 36 factors

so N will have 36 factors that are multiples of 10

Answer: Option B

TestPrepUnlimited · Answer

To end with a 0, we need at least one factor of two and at least one factor of five (to create a factor of 10). Given N, we are allowed to pick up to 3 multiples of two, up to 2 multiples of three, and up to 4 multiples of five.

We need at least one multiple of two, so we can choose from 1, 2, or 3 multiples of twos (3 options).

Similarly, we have 4 choices for fives.

For three we can choose 0, 1, or 2 multiples so that is 3 choices.

Putting everything together we have 3 * 4 * 3 = 36 choices.

Ans: B

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