gmatkkvinu wrote:how can it be solved without using options?
You would have to use modular arithmetic and find an easier number to which 117^(some power) is congruent, modulo 100. The GMAT doesn't expect you to know how to do that. Given that your question only has four answers, though, I'm guessing it's from India's CAT exam, which might expect you to know how. If you're familiar with modular arithmetic, though, it's solvable.
Here's a fairly painful way to solve it (there must be an easier one that I'm not seeing). 117 and 17 have the same remainder (17) when divided by 100. So the question is more easily phrased as "What are the last two digits of 17^(513)?"
Now let's break up 17. 17^3 = 4913, so 17^3 has a remainder of 13 when divided by 100. Since 17^(513) = (17^3)^(171), we know that (17^3)^(171) will have the same remainder as 13^(171).
13^3 ends in 97, so it has a remainder of -3 when divided by 100. (The GMAT doesn't do negative remainders, but they're fine in mod.) Since (13^(171)) = (13^3)^(57), we know that 13^(171) will have the same remainder as (-3)^57. Now we're almost done.
(-3)^57 = 9^28 * (-3)
9^10 has remainder 01 when divided by 100, and 9^4 has remainder 61. So 9^28 = 9^10 * 9^10 * 9^4 * 9^4, which gives us remainder 01 * 01 * 61 * 61, or 3721.
We still have to multiply by that last (-3), so we have (-3) * 3721, or -11163, which has a remainder of 37 when divided by 100. So that's our answer.
Hopefully this illustrates why the answers are what they are!
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EDIT:
I knew there was an easier way!
We can use
Euler's theorem.
Since 17^40 has remainder 1 when divided by 100, we just do
17^(513) = (17^40)^12 * 17^(33)
So the remainder is really that of 17^(33) divided by 100, or of 13^11, or of (-3)^3 * 169, or -27 * 169, or -4563, which has remainder 37 when divided by 100.
This is quite beyond the GMAT, however.
