E.

The units digit of 3^x repeats:

Units digit:
3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1
3^5 = 3

So the pattern is...
3, 9, 7, 1, 3, 9, 7, 1, 3 ...

So the units digit of 3^(8n+3) would be 7. Add 2 to this, and the units digit is 9. The remainder, is 4.

.. another way to do this is to make use of the binomial theorem ..

... 3^(8n+3)+2 = 27*3^8n+2 = 27*9^4n+2 = 27*(10-1)^4n+2 ... now when u expand the expression (10-1)^4n , using the binomial theorem u wuld notice that each term except the last one wuld have 10 in it .. and the last term wuld be 1 .. so when u divide the expression (10-1)^4n by 5 the remainder will be 1 ... so the remainder for 27*(10-1)^4n will be 27 add 2 to it and the remainder will be 29 divide it by 5 and u get the remainder as 4 ...

Now, there is a reason i have illustrated this method....bcoz benys method would work as long as the divisor is 5 or 10 .. but if the divisor is some other number the binomial theorem method will help in solving it ..

PS .. read more about the binomial theorem here https://www.purplemath.com/modules/binomial.htm

Thanks for the alternative method!

I'm not sure if GMAT goes that deeply into math, however. Most of the time that I've seen this type of problem, it's more about recognizing a repeating pattern. Still, good to know.

Thanks.

Yes, I also found a pattern 3,9,7,1... but could not figure if 3^(8n+3) would be 7 (I thought the digit would depend on n as well...) so, no matter what the n would be, we can simply see it as 3^3??

magical cook wrote:Thanks.

Yes, I also found a pattern 3,9,7,1... but could not figure if 3^(8n+3) would be 7 (I thought the digit would depend on n as well...) so, no matter what the n would be, we can simply see it as 3^3??

3^(8n+3) = 3^8n * 3^3 = (3^4)^2n * 27 = 81^2n * 27
Now, any power of 81 will have the units digit of 1. That multiplied by 27 will have a units digit of 7. Add 2 and you get 9. The remainder will be 4.