jlgdr wrote:
Thanks Karishma, anyways what I tried here was prime factorization to see which answer choice could be multiples of 9, that is having the whole sum of digits to be a multiple of 9 and only C and D worked. Now, it seems a bit complicated to decide between the two. I have both even too. Any clues on how to proceed from here? I still haven't used the fact that they need to be divisible by 7 and 8 but I'm not sure how to approach these tests quickly
Kudos for a good reply!
Cheers
J
523abc is divisible by 9 which means 5+2+3+a+b+c is divisible by 9 i.e. 1+a+b+c is divisible by 9.
When you prime factorize each option, you are left only with (D)
A. \(504 = 2^3 * 3^2 * 7\)
Since a, b and c must be single digit numbers, the only way you can assign values to them is 8, 9 and 7. 1+8+9+7 = 25 which is not divisible by 9.
B. \(532 = 2^2*7*19\)
19 cannot be assigned to any of a, b and c and hence this product is not possible.
C. \(210 = 2^2 * 3 * 5 * 7\)
You cannot combine them in any way such that you get three single digit numbers. Note that if 5 and 7 are multiplied by even 2, they will become double digit hence two of a, b and c must take values of 5 and 7. The third cannot be the leftover 12.
D. \(180 = 2^2 * 3^2 * 5\)
There are two options: 4, 9, 5 and 6, 6, 5.
6, 6, 5 satisfies since 1+6+6+5 = 18 divisible by 9
E. \(280 = 2^3 * 5 * 7\)
The digits must be 8, 5 and 7. But 1+8+5+7 is not divisible by 9.
Answer (D)
I find this method to be very time consuming.