Perfect Square Question

[sal2]

K is a positive integer and 225 and 216 are both divisors of k. If k=2^a + 3^b + 5^c, where a, b, and c are positive integers, what is the least possible value for a + b + c

a) 4
b) 5
c) 6
d) 7
e) 8

when you break this one down into prime numbers, for 225 you get 5*5*3*3 and for 216 you get 2*2*2*3*3*3. So, it looks like you will need another 2 and 3 to get the perfect square for a grand total of 2 5s, 6 3s, and 4 2s. Since 12 is not an answer, it looks like you can cut out some of those 3s and 2s. Can someone please explain to me the logic?

[Giorgio]

First of all i think you have a typo , to solve this you have to multiply values in numerator and not add them.

So if we have Values of K= 2^a x 3^b x 5^c

if K is devisable by both 225( 5 , 5, 3,3) and 216 (2,2,2,3,3,3) we can take the least common divisor which will be 5x5x3x3x3x2x2x2 or 5^2 3^3 and 2^3

So the values of a b c , are a =3 b = 3 and c=2 .

we get E) 8

Oops i've just read your comment , to get least common divisor you have to cancel those 2 pairs of 3's , cause they are used in both 225 and 216.

[sal2]

Got it, thanks. I don't know why I had in my mind that it was a perfect square question.