solution-verification | Show that the number $10^n+100^n+1000^n-3$ is divisible by 9, whatever $n$ is a natural number.

Question

the problem

Show that the number $10^n+100^n+1000^n-3$ is divisible by 9, whatever $n$ is a natural number.

my solution

if $n\geq 1$

First I wrote $10^n+ 10^{2n}+10^{3n}-3=10^n[1+10^n(1+10^n)]-3=100...0100...0100...0-3$

        how many times are the 0 repeating                         n-1    n-1     n     

then we get that the form of the numbers is actually $100...0100...0999...97$

        how many times are the 0 repeating       n-1    n   n-1    

The number has the sum of the digits divisible by $9$ so the number is divisible by $9$

The case where $n=0$ can be easily verified

Im not sure if my idea is correct and if I got to the correct form. Hope one of you can help me! Thank you!

Answer 1

Your idea is correct, though It uses a "brutal force" approach. Here is an alternative proof:

Note that $10^n$, $100^n$ and $1000^n$ each leave a remainder of 1 when divided by 9. We denote this by $10^n \equiv 1(\bmod 9)$. For more details about this notation, see Modular Arithmetic. Using some properties of modular arithmetic, we have $$ 10^n+100^n+1000^n-3 \equiv 1+1+1-3\equiv0(\bmod 9) $$ Therefore, it is divisible for 9.

If you prefer not to use this notation, we can express in the following way. There exist integers $a,b$ and $c$ such that $10^n=9a+1$, $100^n=9b+1$, and $1000^n=9c+1$. Therefore: $$ 10^n+100^n+1000^n-3=9(a+b+c)+3-3=9(a+b+c). $$ This shows that the expression is indeed divisible by 9.

Answer 2

Alternative solution:

As, $$10^n=999...9 +1 \ = 9k_n +1$$ Hence $10^n + 100^n +1000^n = 10^n + 10^{2n} + 10^{3n} = 9(k_n + k_{2n} + k_{3n}) + 3$. Which is enough to prove your result.