Bunuel wrote:
azule45 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
A 0
B 1
C 2
D 3
E 4
It should be 3^{8n+3}+2.
The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit:
3^1 --> the units digit is 3;
3^2 --> the units digit is 9;
3^3 --> the units digit is 7;
3^4 --> the units digit is 1;
3^5 --> the units digit is 3 AGAIN;
...
So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... 3^{8n+3} will have the same units digit as 3^3, which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of 3^{8n+3}+2 will be 7+2=9. Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4.
Answer: E.
Check Number theory chapter of Math Book for more: math-number-theory-88376.html
Hope it helps.
I just came to testing the cycle and felt a dead end. I am not getting why are we zeroing in on 3^3 for the first part of the addition in the question