Track / Stage 4 / 63 of 340 #323 of 1964

Problem 323

AMC 12 late, AIME early

Number theory Difficulty 4.7 Prove it South African Mathematics Olympiad Third Round · South Africa

Prove that there are infinitely many terms of the arithmetic sequence $1, 14, 27, 40, \dots, 1+13k, \dots$ which are of the form $222\ldots22$ . In other words a number that is made up using only the digit $2$ .

(Hint: $1001 = 7 \times 11 \times 13$ )

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Note that $221 = 13 \times 17$ and therefore $222 = 13 \times 17 + 1$ is in the arithmetic sequence. Furthermore, $222\,222\,221$ is divisible by $13$ and hence $222\,222\,222$ is in the sequence, since
$222222 = 222 \times 1001 = 222 \times 7 \times 11 \times 13.$
In fact, any number consisting of $6m + 3$ $2$ s will be one more than a multiple of $13$ , since the number consisting of $6m$ $2$ s will be a multiple of $1001$ and hence a multiple of $13$ . Thus there are infinitely many terms in the sequence that only uses the digit $2$ .

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