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by Uva@90 » Mon Oct 14, 2013 3:01 am
[email protected] wrote:What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0
B. 3
C. 2
D. 5
E. None of the above

Ans-B
Shibsriz,
I will give u a clue,
9 to any power goes like this {9,1,9,1,9,1.........}

Try it now.

Regards
uva.
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by Brent@GMATPrepNow » Mon Oct 14, 2013 3:34 am
[email protected] wrote:What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0
B. 3
C. 2
D. 5
E. None of the above

Ans-B
First recognize that a number that's divisible by 6 must be even AND must be divisible by 3.
Also recognize that the product of any number of odd integers will always be odd. So, since 9 is odd, we know that:
9^1 is odd, and it is divisible by 3
9^2 is odd, and it is divisible by 3
9^3 is odd, and it is divisible by 3
9^4 is odd, and it is divisible by 3
.
.
.
9^9 is odd, and it is divisible by 3

So, 9^1 + 9^2 + 9^3 +...+ 9^9 = odd + odd + odd .... + odd = (9)(odd) = ODD
Also, 9^1 + 9^2 + 9^3 +...+ 9^9 is divisible by 3.

So, 9^1 + 9^2 + 9^3 +...+ 9^9 is divisible by 3 and it's odd.
So, if we subtracted 3 (an odd number) from 9^1 + 9^2 + 9^3 +...+ 9^9, we'd get an even number that's divisible by 3.
In other words, 9^1 + 9^2 + 9^3 +...+ 9^9 - 3 is divisible by 6.
So, when we divide 9^1 + 9^2 + 9^3 +...+ 9^9 by 6, we must get a remainder of 3

Answer: B

Cheers,
Brent
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by GMATGuruNY » Mon Oct 14, 2013 4:45 am
What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0
B. 3
C. 2
D. 5
E. None of the above
9¹ + 9² + 9³ + 9� + 9� + 9� + 9� + 9� + 9� = the sum of 9 multiples of 9 = an ODD MULTIPLE OF 9.

Divide odd multiples of 9 by 6 and LOOK FOR A PATTERN.
9/6 = 1 R3.
27/6 = 4 R3.
45/6 = 7 R3.
63/6 = 10 R3.
In every case, the remainder is 3.

The correct answer is B.
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