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

(A) 22
(B) 32
(C) 42
(D) 50
(E) 112

The number of consecutive zeroes at the end of an integer is always going to be the smaller of:
    The number of factors of 2
    The number of factors of 5
For the reasoning you can visualize the following: on a factory floor, two conveyor belts are bringing you prime numbers. One conveyor belt brings you 2s and the other brings you 5s.
Your job is to grab one of each and package the pairs together (factors of 10).
This question is asking how many such pairs you can package together before you run out of 2s or 5s.
125^14 is also known as 5^42, so you have 42 5s coming in.
48^8 is also known as 2^32 * 3*8, so you have 32 2s coming in.
You can package together a total of 32 pairs before you run out of 2s.
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=> \(125^{14} = (5^3)^{14}\)

=> \(48^{8} = (2^4)^8 * (3)^8\)

=> \(125^{14} * 48^{8} = (5^3)^{14} * (2^4)^8 * (3)^8 = (5)^{42} *(2)^{32} * (3)^8\)

=> \((5)^{42} *(2)^{32} * (3)^8 = (2 * 5)^{32} * (5)^{10} * (3)^8\)

=> \((5)^{42} *(2)^{32} * (3)^8 = (10)^{32} * (5)^{10} * (3)^8\)

=> (10)^{32} = ..32 zeroes

Answer B
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