If \((125^{14})(48^8)\) is written out as an integer, how many consecutive zeroes will that integer have at the end?

To find the number of consecutive numbers we need to find what is the power of 10 in the given number.
And to find the power of 10 we need to find out how many 5's and 2's are there which can get multiplied to give us 10

\(125^{14}*48^8\)
125 can be written as \(5^3\)
=> \((5^3)^{14}*48^8\) (and 48 can be written as 16*3 )
=> \(5^{42}*(16*3)^8\) (16 can be written as \(2^4\))
=> \(5^{42}*(2^4*3)^8\)
=> \(5^{42}*2^{4*8}*3^8\)
=> \(5^{42}*2^{32}*3^8\)
=> \(5^{10}*5^{32}*2^{32}*3^8\)
=> \(5^{10}*10^{32}*3^8\)

So, we can have 32 consecutive 0's in the end as we have \(10^{32}\) as the maximum power of 10 in the number.

So, Answer will be B
Hope it helps!

Watch the following video to MASTER Exponents

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.