### Code to Zero

Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.

### N000ughty Thoughts

How many noughts are at the end of these giant numbers?

### Mod 3

Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

# Obviously?

##### Stage: 4 and 5 Challenge Level:

Murat from Turkey sent the following solution.

If $g(n)=(1+8^n-3^n)$ is divisible by 6, then $(1+8^n-3^n-6^n)$ is also divisble by 6. It can be verified that

$$g(n)-73g(n-2)+576g(n-4)=504$$

for all positive integers $n> 3$. Since $g(1)$ and $g(3)$ are divisible by 6, it follows that $g(5)$ is also. By induction, it can be shown that for all odd $n$, $g(n)$ is divisible by 6. Since $g(2)$ and $g(4)$ are not divisible by 6, this is not the case for even $n$.

An alternative approach is to use the facts that powers of odd numbers are always odd; powers of even numbers are always even; also the difference of two odd numbers is even. Hence $N=1^n+8^n-3^n-6^n$ is even (odd + even - odd - even).

It remains to decide whether or not $N$ is divisible by 3.

$N \equiv 1 + (-1)^n - 0 - 0$ (mod 3)

This shows that $N \equiv 0$ (mod 3) if $n$ is odd and hence $N$ will be divisible by 6 for all odd values of $n$. However, $N \equiv 2$ (mod 3) if $n$ is even and so $N$ cannot be divisible by 6 for even values of $n$.