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Re: If a, b and c are any three single digit numbers from 1 to 9, both [#permalink]

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01 Aug 2017, 02:00

I say E.

The sum must be equal to 9 or 18, in order for (a+b+c) to be divisible by 9.

A and B are clearly not sufficient, as we are not given infos on the third number (c and a respectively). Thus, D is out aswell.

C is appealing. Try to pick some random numbers and you will most likely come up with three values whose sum is different from 9 and 18. So THERE ARE combinations of a+b+c resulting in a number non-divisible by 9. Are there any divisible by 9?

Personally, I solved it by assuming b (the common number in both A and B) was 9, and looked for combinations of a+b=9 (so that 9+9=18). The most immediate is a=5 c=4. Notice that this is rather quick, since 6ab and 4bc, having b=9, will 100% be multiples of 9, thus divisible by 9.

If a, b and c are any three single digit numbers from 1 to 9, both [#permalink]

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03 Aug 2017, 11:34

1

This post was BOOKMARKED

If a, b and c are any three single digit numbers from 1 to 9, both inclusive, is (a + b + c) divisible by 9?

(1) The number 2ab3 is divisible by 9. (2) The number 4bc1 is divisible by 9.

now as per (1); a+b=4 or 13 also as per (2); b+c=4 or 13;

BUT if (a+b)=4 then (b+c)= 13 as it can't be 4 or if (b+c)=4 then (a+b)= 13 as it can't be 4

so; one of (a+b) and (b+c) is 4 and other is 13...

So, we can say that (a+2b+c)=17 for sure;

Now; since (a+2b+c)=17...then (a+b+c) can only have its value as 9 to be able to be perfectly divided by 9.. and for that to happen.. value of b must be 8...but it can't be possible as one of (a+b) and (b+c) is 4...So b is surely less than 4..

Therefore...combining (1) and (2)...we can say that a+b+c is not divisible by 9... Hence option "C" is correct
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