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.
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in
10 000! and 100 000! or even 1 000 000!
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
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
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 $n$.