Bunuel
What is the greatest power that 5 can be raised to so that the resulting number is a factor of 15!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Looking at 15! (pronounced 15 factorial), we probably won’t spend the time on test day to calculate that it can be rewritten as 1,307,674,368,000. However I already knew that it would have to end in exactly three 0’s. Why? Because 0 is only formed by multiplying 2 by 5. Thus every zero will indicate the presence of both a 5 and 2. The three 0’s are the result of multiplying 2’s and 5’s in the prime factors of these numbers. Unsurprisingly, there will be a lot of 2’s in this product, but only a handful of 5’s.
How many 5’s exactly? Well 5 is a prime number, so it won’t be formed by the product of two different factors. It will only appear in numbers that have a five in them, or the numbers that end in 5 or 0. In 15! there are three numbers that will have a 5 in them: 5, 10 and 15. Thus we know that this number is the product of 53 as well as several other prime numbers.
Solving this problem does not require us to take the full prime factorization of 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, but you certainly could do it if you wanted to. The answer is 211*36*53*72*111*131. (Fun fact, this number will have exactly 4032 factors… collect them all!). The only number we are interested in is that 53, which means that we can raise 5 to the third power and still get a factor of 15 factorial. Thus, 54 will not qualify.
Answer choice B is correct.