suhaschan wrote:
If n is a positive integer, what is the remainder when 3^"8n+3" + 2 is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4
You should know the concept of cyclicity. Powers of 3 have a cyclicity of 4 i.e.
3^1 =
33^2 =
93^3 = 2
73^4 = 8
13^5 = 24
33^6 = 72
9Look at the last digits of powers: 3, 9, 7, 1, 3, 9, 7, 1, 3, 9...
So whenever we have a power of 3, we remove groups of 4 out of it i.e. divide it by 4 and whatever is the remainder tells us the last digit.
So say we have 3^13, we divide 13 by 4 and are left with 1 remainder. So this is the first term of the next cycle and will end with 3.
Say we have 3^39, we divide 39 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7.
If you have \(3^{8n+3}\), we divide 8n+3 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7.
Last term of \(3^{8n+3}+2\) will be 7+2 = 9. When you divide a number that ends with 9 by 5, you get 4 remainder. If you are not sure why this is so, check out the first part of the question discussed in this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/03 ... emainders/Answer (E)
Or since you know that cyclicity of 3 is 4, just put n = 0 to get 3^3 + 2 = 29. When you divide 29 by 5, you get remainder 4.
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