Prove that is divisible by 7 (condition)

Question

Determine the digit $ m $ tal, $ 888 … 8m9 … 999 $ (a number of times $8$ and $9$ repeat $50$ times) is divisible by $ 7 $

We have given a $101$-digit natural number divisible by $7$ defined as

$(1) \;\; N_m = \underbrace{888 \ldots 88}_{50} m \underbrace{999 \ldots 99}_{50}$.

Assume $m \neq 9$. Then by $(1)$

$(2) \;\; N_m + 1 $ We have given a 101-digit natural number divisible by $7$ defined as

$(1) \;\; N_m = \underbrace{888 \ldots 88}_{50} m \underbrace{999 \ldots 99}_{50}$.

Assume $m \neq 9$. Then by (1)

$(2) \;\; N_m + 1 = \underbrace{888 \ldots 88}_{50} (m+1) \underbrace{000 \ldots 00}_{50} $

I think this way is very technical, I don't know exactly …

Answer 1

Note that $111\,111=7\times15\,873$. So, since $6\mid48$, the numbers$$\overbrace{88\ldots88}^{48\text{ times}}\ \text{ and }\ \overbrace{99\ldots88}^{48\text{ times}}$$are multiples of $7$. And your number is the sum of three numbers:

• $\displaystyle\overbrace{88\ldots88}^{48\text{ times}}\times10^{53}$;
• $\displaystyle88m99\times10^{48}$;
• $\displaystyle\overbrace{99\ldots99}^{48\text{ times}}$.

The first and the third of these numbers are multiples of $7$. And, since $7$ and $10$ are coprime, $7\mid88m99\times10^{48}$ if and only if $7\mid88m99$. Besides, this assertion holds if and only if $7\mid11m22$. With a little effort, you can see that this happens if and only ff $m=5$.