# Quantitative aptitude

I'm not very good at maths, but I need help with a question. This one appeared in an aptitude test.

Consider the sum: $ABC + DEF + GHI = JJJ$ .If different letters represent different digits, and there are no leading zeros, what does $J$ represent?

The solution says that $J=9$, but I don't understand how.

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The point is that there are ten letters, and that each one represent a different cipher. So, one point is to try all possibilities, and other to consider that A, D, G and J are no zero because leading coefficients are not zero and some equation that comes from the sum. –  Josué Tonelli-Cueto Sep 4 '11 at 18:57
Thanks for commenting. Yes I got the bit about A, D, G and J being non zero. But how do I go about testing for all possible cases here in a time limited environment (an aptitude test) for problems of this nature. As I said, I'm not really a math expert, but I can grasp the logic. Do you have any way to find this quickly - a "trick" ? –  yati sagade Sep 4 '11 at 19:08
ABC+DEF+GHI=JJJ 140+273+586=999 –  Mukesh Reddy Jul 5 '12 at 12:48
@Mukesh: The question didn't ask for a solution, but rather why $J=9$. –  Asaf Karagila Jul 5 '12 at 14:16
@Asaf Karagila: by trial and error i got this –  Mukesh Reddy Jul 6 '12 at 5:42

Consider the sum modulo 9 ("casting out nines") so that a number becomes equal to the sum of its digits.

Then $A+B+C+D+E+F+G+H+I+J = J+J+J+J$ with the first sum being (0+1+2+3+4+5+6+7+8+9)=45=0 modulo 9. This forces $4J$ and thus $J$ to be divisible by 9.

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+1,this is a useful approach,and it also proves that $J=9$ is the only solution for this problem. –  Quixotic Sep 4 '11 at 19:41
That was elegant, though I had to read it 4 times :) –  yati sagade Sep 4 '11 at 19:44
It seems like a difficult problem for a general-purpose aptitude test. What test did it come from? –  zyx Sep 4 '11 at 19:50
@yati:It doesn't matter as long as you understand it correctly:) –  Quixotic Sep 4 '11 at 19:50
@zyx Accenture :) –  yati sagade Sep 4 '11 at 19:51
Using this same idea for the option $9$, two possible partitions are $\{4,3,2\}\{7,1,1\}$ notice here there is two $1$ in the second partition,so for atleast one row sum we need a $19$ to carry up that deficit $1$ implicitly,hence $\{2,8,9\}$.
Hence we can conclude one possible solution as: $$402 +318+279 = 999$$ You can see this problem could have multiple possible solution for the sum to be all $9$'s.