If 5^k is a factor of the product of odd integers from 99 to 199, what

Easiest way is to use AP

(Don't be intimidated by the long answer given below. Ive overexplained the solution)

The question states that odd no.s between 99 and 199 are multiplied ie. 99 x 101 x 103......197x199

It also states that 5^k is a factor ie. completely divisible to the above product. Hence all 5s in the denominator must be eliminated.

Now any multiple of 5 must end in either 5 or 0.

But the above product has only odd no.s ie there are no no.s ending with 0.

Hence, the no.s we are looking for are 105,115,125.....,195

last term (l) = 195
First term(a) = 105
Common difference(d)=10

Now, l=a+(n-1)d
195=105+(n-1)10
n=10

Hence, k must also equal to 10. If 5^10 was the denominator, it would get eliminated completely.

But a few no.s in the product can be divided by 5 even further.

125/5=25
25/5=5
5/5=1

Since we had already counted the first 5, we get two more 5s.
Now, k=12.

Again, 175 can be divided more than once.
175/5=35
35/5=7

Since we already counted the first 5, we get another 5.

Hence, we get k=10 + 2 + 1 = 13.

Hence, the answer B

Odd multiples of 5 - 105,115,125,135,....,195. Total=10 (I have used the formula for AP).
Odd multiples of 25 - 125,175. Total=2. These two terms are also involved in the above series.
Remove 125 and 175 from series 1 for ease. So total terms in series 1=8.
Now in 1st series, each term has one 5. In the second series, 125 has 3 5s and 175 has 2 5s. Therefore total 5s are 8+3+2=13(B).