Product with Repdigit can be Split into Parts which Add to Repdigit

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Theorem

Let $n$ be a positive integer with $d_1$ digits.

Let $m$ be a repdigit number with $d_2$ digits such that $d_2 > d_1$.

Let $r$ consist of the result when the rightmost $d_2$ digits of $m n$ is cut off and added to the remaining left hand portion.


Then $r$ is a repdigit number.


Proof

Let $b > 1$ be the base we are working on.

Let $m = \sqbrk {aaa \dots a}_b$.

Let $R = \dfrac m a = \sqbrk {111 \dots 1}_b = \dfrac {b^{d_2} - 1} {b - 1}$.

Let the rightmost $d_2$ digits of $m n$ be $y$ and the remaining left hand portion be $x$.

Then we have:

\(\ds 0\) \(\equiv\) \(\ds m n\) \(\ds \pmod R\)
\(\ds \) \(\equiv\) \(\ds b^{d_2} x + y\) \(\ds \pmod R\)
\(\ds \) \(\equiv\) \(\ds \paren {\paren {b - 1} R + 1} x + y\) \(\ds \pmod R\)
\(\ds \) \(\equiv\) \(\ds x + y\) \(\ds \pmod R\)
\(\ds \) \(\equiv\) \(\ds r\) \(\ds \pmod R\)

so $r$ is divisible by $R$.


It remains to show that:

$r \le b^{d_2} - 1 = \paren {b - 1} R$

because $\paren {b - 1} R$ is the largest $d_2$-digit integer.

This condition forces $r$ to have $d_2$ digits, so it remains a repdigit number.

Since $r$ is divisible by $R$, this is equivalent to:

$r < b R$

We have:

\(\ds r\) \(=\) \(\ds x + y\)
\(\ds \) \(<\) \(\ds b^{d_1} + \paren {b - 1} R\)
\(\ds \) \(\le\) \(\ds b^{d_2 - 1} + \paren {b - 1} R\) because $d_2 > d_1$
\(\ds \) \(\le\) \(\ds R + \paren {b - 1} R\) because $R$ has $d_2$ digits
\(\ds \) \(=\) \(\ds b R\)

Hence the result.

$\blacksquare$


Examples

Product of $894$ with $22 \, 222$

\(\ds 894 \times 22 \, 222\) \(=\) \(\ds 19 \, 866 \, 468\)
\(\ds 198 + 66 \, 468\) \(=\) \(\ds 66 \, 666\)


Sources

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