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Bunuel
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well, it's easier if you few digits with their powers up till 10 and first 10 factorials and their values.
or let me tell you a simple way
3^7 = 3.3.3.3.3.3.3
so rather than actually multiplying 3 with itself 7 times, you can club it up = 9.9.9.3 = 729*3
i hope this helps.
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n = 4q + 3
possible values of n are 3,7,11,15,19...

The cyclicity of powers of 3 is: 3,9,7,1

By looking at all the possible values of n, you can quickly see that they will always end up being in the third place of the cyclicity of powers of 3 (as you increase every possible value of n by 4 every time)

Therefore, the units digit will always be a 7

therefore the remainder of 3^n + 2 will always be 9/5 = 4
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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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