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If a and b are positive integers, is (10^a) + b divisible by

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If a and b are positive integers, is (10^a) + b divisible by [#permalink]

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New post 29 Jul 2009, 16:50
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If a and b are positive integers, is (10^a) + b divisible by 3?

1. b/2 is an odd integer.
2. the remainder of b/10 is b

This is from the Gmatclub tests. The answer is E. But no explanation was provided for Statement 2.

Is S2 possible only when b=0?
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Re: Number Property - DS Question [#permalink]

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New post 29 Jul 2009, 18:22
Ans is E.

To answe this question, we need to remember that, for any integer to be divisible by 3, the sum of the digits has to be 3 or multiple of 3.

now the given expression = (10^n) + b
whatever be the value of n, the sum of the digits of (10^n) is always 1.
so, the value of b will determine whether (10^n) is divisible by 3.

statement 1 says: b/2 is odd.
which means b is even (even*odd=even). but this is not sufficient.
for example, if b=2, then [(10^n) +2] will have the sum of its digits equal to 3 and hence will be divisible by 3.
but, if b=4, then the sum of digits of [(10^n) +4] equals 5 which is not divisible by 3.
hence statement 1 is not sufficient.

statement 2 says, remainder of b/10 = b, which means b is less than 10.
again, by the above examples we can say this statement also is not sufficient.

combining the two statements we get b is positive even integer less than 10. this also is not sufficient by the same examples.

hence E.
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Re: Number Property - DS Question [#permalink]

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New post 29 Jul 2009, 21:48
If a and b are positive integers, is (10^a) + b divisible by 3?

1. b/2 is an odd integer.
2. the remainder of b/10 is b


10^a will always give remainder of 1 .......b ??

from 1

b is an even intiger (2/3..2, 4/3...1, 6/3...0, 8/3...2,..etc)...insuff

from 2
1<= b<=9............insuff

both E
Re: Number Property - DS Question   [#permalink] 29 Jul 2009, 21:48
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