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From Kaushik Srinivasan
Find all $n$ that $3^n + 7^n$ is divisible by 10\par We notice that positive powers of 3 end in 3, 9, 7, 1 (i.e $3^1, 3^2, 3^3, 3^4$) and these digits repeat themselves.
Also the powers of 7 end in 7, 9, 3, 1 repectively.
Hence when the odd powers of 3 and 7 are added we get a zero as the last digit thus divisible by 10.\par Hence when $n$ is odd $3^n + 7^n$ is divisible by 10.
Part solutions were received from a number of our members. Ngoc Tran, Clement Goh (River Valley High School) and Andrei Lazanu (School 205 Bucharest). You were all able to spot patterns but no one explained why or how you might predict what will happen. For example why does 3 4n always end in 1?
Power Crazy
Age 11 to 14
Challenge Level 
From Kaushik Srinivasan
Find all $n$ that $3^n + 7^n$ is divisible by 10\par We notice that positive powers of 3 end in 3, 9, 7, 1 (i.e $3^1, 3^2, 3^3, 3^4$) and these digits repeat themselves.
Also the powers of 7 end in 7, 9, 3, 1 repectively.
Hence when the odd powers of 3 and 7 are added we get a zero as the last digit thus divisible by 10.\par Hence when $n$ is odd $3^n + 7^n$ is divisible by 10.
Part solutions were received from a number of our members. Ngoc Tran, Clement Goh (River Valley High School) and Andrei Lazanu (School 205 Bucharest). You were all able to spot patterns but no one explained why or how you might predict what will happen. For example why does 3 4n always end in 1?
A spreadsheet was a really useful tool for looking for patterns by quickly calculating powers of each of the numbers and then looking for patterns, but not for explaining why. Why is the power always of the form 2k+1 of 4k+2 except for 5 n + 5 n ?
Here are some combinations (n and k are integers):
| 1 n + 1 n | never (why?) |
| 1 n + 2 n | never (why?) |
| 1 n + 3 n | when n = 4k+2 (why?) |
| 1 n + 4 n | never |
| 1 n + 5 n | ever |
| 1 n + 6 n | never |
| 1 n + 7 n | when n = 4k+2 |
| 1 n + 8 n | never |
| 1 n + 9 n | when n = 2k+1 |
| 2 n + 2 n | never |
| 2 n + 3 n | never |
| 2 n + 4 n | when n = 4k+2 |
| 2 n + 5 n | never |
| 2 n + 6 n | when n = 4k+2 |
| 2 n + 7 n | never |
| 2 n + 8 n | when n = 2k+1 |
| 2 n + 9 n | never |
| 3 n + 3 n | never |
| 3 n + 4 n | never |
| 3 n + 5 n | never |
| 3 n + 6 n | never |
| 3 n + 7 n | when n = 2k+1 |
| 3 n + 8 n | never |
| 3 n + 9 n | when n = 4k+2 |
| 4 n + 4 n | never |
| 4 n + 5 n | never |
| 4 n + 6 n | when n = 2k+1 |
| 4 n + 7 n | never |
| 4 n + 8 n | when n = 4k+2 |
| 4 n + 9 n | never |
| 5 n + 5 n | when n=k |
| 5 n + 6 n | never |
| 5 n + 7 n | never |
| 5 n + 8 n | never |
| 5 n + 9 n | never |
| 6 n + 6 n | never |
| 6 n + 7 n | never |
| 6 n + 8 n | when n = 4k+2 |
| 6 n + 9 n | never |
| 7 n + 7 n | never |
| 7 n + 8 n | never |
| 7 n + 9 n | never |
| 7 n + 9 n | when n = 4k+2 |
| 8 n + 8 n | never |
| 8 n + 9 n | never |
| 9 n + 9 n | never |
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Adding All Nine
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
