If a and b are positive integers, what is the remainder when : GMAT Data Sufficiency (DS)
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# If a and b are positive integers, what is the remainder when

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If a and b are positive integers, what is the remainder when [#permalink]

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26 Feb 2012, 15:30
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If a and b are positive integers, what is the remainder when $$4^{2a+1+b}$$ is divided by 10?

(1) a = 1
(2) b = 2

[Reveal] Spoiler:
Ok - this is how I am trying to solve this.

Statement 1

a = 1. Does not tell anything about b --therefore is insufficient on its own to answer the question.

Statement 2

b = 2

2a + 1 + b becomes

2a (even) + 1 (odd) + b (even) = ODD. So the exponent to 4 is ODD. So I understand that if we put 3, 5 etc I get the remainder 4, but why can't I put exponent as 1 as 1 is ODD too. Can you please help?
[Reveal] Spoiler: OA

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Re: Remainder when divided by 10 [#permalink]

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26 Feb 2012, 15:39
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powers of 4 go like this:

The unit place is:
1 = 4
2 = 6
3 = 4
4 = 6

So all even exponents have 6 in unit place, and all off exponents have 4 in unit place. To solve the problem we need to find whether the 2a + 1 + b is even or odd

As a is +ve integer, 2a is always even. 2a + 1 will be odd. Now to determine whether (2a + 1 + b) is even or odd, we need to know only b.

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Re: Remainder when divided by 10 [#permalink]

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26 Feb 2012, 15:58
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If a and b are positive integers, what is the remainder when $$4^{2a+1+b}$$ is divided by 10?

This is a classic "C trap" question: "C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together. When you see such question you should be extremely cautious when choosing C for an answer.

Back to the question: 4 in positive integer power can have only 2 last digits: 4, when the power is odd or 6 when the power is even. Hence, to get the remainder of 4^x/10 we should know whether the power is odd or even: if it's odd the remainder will be 4 and if it's even the remainder will be 6.

(1) a = 1 --> $$4^{2a+1+b}=4^{3+b}$$ depending on b the power can be even or odd. Not sufficient.

(2) b = 2 --> $$4^{2a+1+b}=4^{2a+3}=4^{even+odd}=4^{odd}$$ --> the remainder upon division of $$4^{odd}$$ by 10 is 4. Sufficient.

enigma123 wrote:
2a (even) + 1 (odd) + b (even) = ODD. So the exponent to 4 is ODD. So I understand that if we put 3, 5 etc I get the remainder 4, but why can't I put exponent as 1 as 1 is ODD too. Can you please help?

The power of 4 is $$2a+3$$ and since $$a$$ is a positive integer then the lowest value of $$2a+3$$ is 5, for $$a=1$$. Next, even if the power were 1 then 4^1=4 and the remainder upon division of 4 by 10 would still be 4.

Hope it's clear.
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Re: If a and b are positive integers, what is the remainder when [#permalink]

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11 Aug 2013, 08:20
REM(4^(2a+1+b))/10

Means we have to find last digit of the expression.So rephrasing the question

What is the last digit of 4^(2a+1+b)

(1).
a=1

Break the expression as 4^2a * 4^1 * 4^b.

b is unknown hence INSUFFICIENT

(2).

b=2

4^2a * 4^1 * 4^b.

If you can observe the expression 4^2a, you will see that this will always give last digit as '6' you can try out numbers if you want.

So knowing the expression and value of b last digit can be calculated and hence the remainder can also be calculated.

Hence (B) it is !!
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Re: If a and b are positive integers, what is the remainder when [#permalink]

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30 Aug 2014, 04:22
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02 Sep 2015, 09:11
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If a and b are positive integers, what is the remainder when [#permalink]

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03 Sep 2015, 03:48
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

If a and b are positive integers, what is the remainder when 4 2a+1+b is divided by 10?

(1) a = 1
(2) b = 2

Transforming the original condition and the question, 4^(2a+1+b)=(4^2a)(4^(1+b))=(16^a)(4^(1+b))=(.......6)(4^(1+b)), because the first digit is always 6 when it's multiplied by 6. Since b is all we need to know, the answer is B.
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If a and b are positive integers, what is the remainder when   [#permalink] 03 Sep 2015, 03:48
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