Carcass wrote:
k is a positive integer and 225 and 216 are both divisors of k. If k−2a×3b×5c, where a, b and are positive integers, what is the least possible value of a+b+c?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Kudos for the right answer and explanation
225=52∗32 and
216=23∗33Since 225 and 216 are both divisors of
k, it must be that
k is a multiple of the LCM of 225 and 216
=>
k = Multiple of
52∗33∗23Since
k=2a×3b×5c, and we need the minimum values of
a,b,c, we take:
=>
k =
52∗33∗23 =
5c∗3b∗2a=>
a=3,
b=3,
c=2=>
a+b+c=8Answer E