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

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

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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.
[Reveal] Spoiler: OA

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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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Bunuel wrote:
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.



stat1: a + b can be 4, 13, 22 etc... no info on c....not suff

stat2: b + c can be 4, 13, 22 etc... no info on a...not suff

both combined...not suff,,

ans E

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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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New post 31 Jul 2017, 13:49
Bunuel wrote:
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.




a+b = 4; 2,2/ 5,8/ 8, 5 ...
b+c = 4; 2,2/ 5,8/ 8, 5 ...

Combining
a,b,c = 2,2,2 not divisible
a,b,c = 5,8,5 divisible

Answer: E

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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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New post 01 Aug 2017, 00:58
Bunuel wrote:
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.


My answer was C because combine together still not divisible..
can I do this?
Please tell me what is wrong with my logic.. Thank you

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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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New post 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.

Answer must be E then.

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

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New post 03 Aug 2017, 11:34
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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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If a, b and c are any three single digit numbers from 1 to 9, both   [#permalink] 03 Aug 2017, 11:34
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