How many positive integers k are there such that 100k is a factor of (

For any number, N broken into prime factors and their powers in the form \(N = a^p * b^q*c^r\) and so on, then the total number of factors n = (p + 1) * (q + 1) * (r + 1) and so on..

We want to know, how many values of k are possible, if 100k is a factor of the given number i.e we need to find out how many factors of \((2^2)(3)(5^3)\) end with 2 zeros.

To get a number ending with a zero, we need a 2 and a 5, and for 2 zeroes, we need 2 twos and 2 fives.

Taking out 2 two and 2 fives, we get \(2^2 * 5^2[3^1 * 5^1]\)

Now by the counting rule for factors, we count the number of factors for the numbers in the bracket i.e \(3^1 * 5^1\)

The number of factors = (1 + 1) * (1 + 1) = 4


Option E

Arun Kumar

I don't understand the highlighted portion, I know the counting rule to count factors, but not understanding why we're counting the remaining 5 and 3 factors.

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