Let n ∈ ℕ. If the sum of the digits of n is equal to the sum of the digits of 5n, then prove that 9|n.

Question

I know how to test the divisibility of a number by 9, but only if I am given what n is. How would I set this problem up?

Answer 1

In general, for 2 nonnegative integers $\overline{a_1\ldots a_n}$ and $\overline{b_1\ldots b_m}$, we have $$ \overline{a_1\ldots a_n}-\overline{b_1\ldots b_m}=(\sum a_i)-(\sum b_j)+9A $$ for some integer $A$. Thus, your premise gives you $9|4n$. What does this say about $n$?

Answer 2

The sum of the digits of $5n$ is equal to the sum of the digits of $n$, hence we can say $n\equiv 5n (\mod 9)\rightarrow 9|(5n-n)\rightarrow 9|(4n)\rightarrow 9|n$

And in the above solution we have used from this fact that $\gcd{(4,9)}=1$