digits

[gmatusa2010]

If a and b are positive integers, what is the remainder when
9^(2a+1+b) is divided by 10?
(1) a = 3
(2) b is odd.

[Anurag@Gurome]

If a and b are positive integers, what is the remainder when
9^(2a+1+b) is divided by 10?
(1) a = 3
(2) b is odd.

Problem is asking for units digit of 9^(2a + 1 + b).
Now if the power of 9 is odd => Unit's digit = 9
And if the power of 9 is even => Unit's digit = 1

Thus we can determine the unit's digit of 9^(2a + 1 + b), if we know whether (2a + 1 + b) is odd or even.

Statement 1: a = 3
(2a + 1 + b) = (b + 7)
We don't whether b is odd or even.

Not sufficient

Statement 2: b is odd
(2a + 1 + b) = (Even + 1 + Odd) = Even

Sufficient

The correct answer is B.

[gmatusa2010]

Is there anyway to do this with base 3? 3^(2(2a+1+b))?

[Anurag@Gurome]

Is there anyway to do this with base 3? 3^(2(2a+1+b))?

Yes, we can do that also.

9^(2a + 1 + b) = 3^(2(2a + 1 + b)) = 3^(4a + 2b + 2)

Now, the units digit of powers of 3 are: 3, 9, 7, 1, 3, 9...
Thus unit's digit of 3^(4a + 2b + 2) = unit's digit of 3^(2b + 2)

Hence it doesn't depend upon a => Statement 1 is NOT sufficient.

When b is odd, simply put b = 1 and we have 3^4, whose unit's digit is 1.

Or, b can be expressed as (2n + 1) for some non-negative integer n.
Thus, (2b + 2) = 2*(2n + 1) + 2 = 4n + 4 = 4(n + 1)
Thus, unit's digit of 3^(2b + 2) = unit's digit of 3^(Some multiple of 4) = 1

Statement 2 is sufficient