Consecutive Triple of Repeated Digit-Products

Theorem

The triplet of integers $281, 282, 283$ have the property that if their digits are multiplied, and the process repeated on the result until only $1$ digit remains, that final digit is the same for all three, that is, $6$.

There does not exist an set of four consecutive integers which also all end up at the same single digit.

Proof

\(\ds 281: \ \ \)

\(\ds 2 \times 8 \times 1\)

\(=\)

\(\ds 16\)

\(\ds 1 \times 6\)

\(=\)

\(\ds 6\)

\(\ds 282: \ \ \)

\(\ds 2 \times 8 \times 2\)

\(=\)

\(\ds 32\)

\(\ds 3 \times 2\)

\(=\)

\(\ds 6\)

\(\ds 283: \ \ \)

\(\ds 2 \times 8 \times 3\)

\(=\)

\(\ds 48\)

\(\ds 4 \times 8\)

\(=\)

\(\ds 32\)

\(\ds 3 \times 2\)

\(=\)

\(\ds 6\)

$\blacksquare$

Historical Note

This result was attributed to Douglas J. Lanska and Charles Ashbacher, according to David Wells, who sourced the result from Journal of Recreational Mathematics, volume $22$ page $70$.

It has not so far proved possible to corroborate this.

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $281$