Kaprekar's Symmetry

Theorem

Let $n$ be a Kaprekar number with $D$ digits.

Then $10^D - n$ is also a Kaprekar number.

Examples

\(\ds 1 + 9\)

\(=\)

\(\ds 10\)

$1$st and $2$nd Kaprekar numbers

\(\ds 45 + 55\)

\(=\)

\(\ds 100\)

$3$rd and $4$th Kaprekar numbers

\(\ds 297 + 703\)

\(=\)

\(\ds 1000\)

$6$th and $7$th Kaprekar numbers

\(\ds 2223 + 7777\)

\(=\)

\(\ds 10 \, 000\)

$9$th and $16$th Kaprekar numbers

\(\ds 2728 + 7272\)

\(=\)

\(\ds 10 \, 000\)

$10$th and $15$th Kaprekar numbers

\(\ds 04879 + 95 \, 121\)

\(=\)

\(\ds 100 \, 000\)

$11$th and $23$rd Kaprekar numbers

\(\ds 4950 + 5050\)

\(=\)

\(\ds 10 \, 000\)

$12$th and $13$th Kaprekar numbers

Proof

Since $n$ is a Kaprekar number of $D$ digits:

$\begin {cases} n^2 = a \times 10^D + b \\ n = a + b \end {cases}$

for some positive integers $a$ and $b$, $b < 10^D$.

Hence:

\(\ds \paren {10^D - n}^2\)

\(=\)

\(\ds 10^{2 D} - 2 n \times 10^D + n^2\)

\(\ds \)

\(=\)

\(\ds 10^{2 D} - 2 \paren {a + b} 10^D + a \times 10^D + b\)

\(\ds \)

\(=\)

\(\ds 10^D \paren {10^D - a - 2 b} + b\)

and we have:

$\paren {10^D - a - 2 b} + b = 10^D - a - b = 10^D - n$

Finally we check that $10^D - a - 2 b \ge 0$:

Aiming for a contradiction, suppose $10^D - a - 2 b \le -1$.

Then:

\(\ds \paren {10^D - n}^2\)

\(=\)

\(\ds 10^D \paren {10^D - a - 2 b} + b\)

\(\ds \)

\(\le\)

\(\ds -10^D + b\)

\(\ds \)

\(<\)

\(\ds 0\)

but squares are positive, a contradiction.

Hence $10^D - n$ is also a Kaprekar number of $D$ digits.

$\blacksquare$

Source of Name

This entry was named for Dattathreya Ramchandra Kaprekar.

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $297$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $297$