remainder theorem problem

Power | Result | Remainder for 32 |
1 | 17 | 17 |
2 | 289 | 1 |
3 | 4913 | 17 |
4 | 83521 | 1 |
5 | 1419857 | 17 |

So the answer should be 4

Another way is:
17 = (0*32)+17
17^2 = 289= 9*32 + 1 (A)
17^3 = 17^2 * 17 = 32*9*17+17 (B)
17^4 = B * 17 = 32*9*17*17+A= 32(9*17*17+9)+1
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Hi Priyanshi,

This question doesn't require this much of working. Based on the facts

Fact : Remainder will only be based on remainders of previous step

which means 17^3 will leave remainder [Rem(17^2) x Rem (17)] / 32

Step 1: 17 / 32 = Rem(17)
Step 2: 17^2 / 32 = 289 / 32 = Rem(1)
Step 3: 17^3 / 32 = (17)x(1) / 32 = Rem(17)

The repetition of remainders has started. For every odd power of 17 the remainder is 17 and for every even power of 17, the remainder is 1

Therefore 17^2011 will have remainder 17 when divided by 32

P.S. A few GMAT test takers in past have reported question based on this principle. Howeven as Brent says we must have 5 options to qualify the question to be a GMAT Quant question

17 ^ 2010 * 17
= ((16 + 1)^2)^1005 * 17
= ( 16^2 + 2 * 16 + 1)^1005 * 17

remainder: 1 * 17 = 17