Competition problem — Number theory

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Number theory Difficulty 7.0 National Olympiad Prove it Japan

For positive 2-digit integers $x$ and $y$ , the ten's digit of $x$ equals the one's digit of $y$ and the ten's digit of $y$ equals the one's digit of $x$ . Let $P$ be the product of $x$ and $y$ . Suppose $P$ is a 4-digit number, and suppose that the 2-digit number given by the last (i.e., the bottom) 2 digits of $P$ is 23 greater than the 2-digit number given by the first (i.e., the top) 2 digits of $P$ . Determine the value of $P$ .

Solution

By the given assumptions of the problem, we can write
$x = 10a + b, \quad y = 10b + a, \quad P = xy = 100c + (c + 23) = 101c + 23,$
where $a, b, c$ are integers satisfying $1 \le a, b \le 9$ and $c$ has 2-digits. We then obtain from
$101c + 23 = P = (10a + b)(10b + a) = 101ab + 10(a^2 + b^2),$
the fact that $10(a^2 + b^2) \equiv 23 \pmod{101}$ holds. Consequently, we have
$a^2 + b^2 \equiv -100(a^2 + b^2) \equiv -10 \cdot 23 = -230 \equiv 73 \pmod{101}.$
From $a^2 + b^2 \equiv 73 \pmod{101}$ and $2 = 1^2 + 1^2 \le a^2 + b^2 \le 9^2 + 9^2 \le 162$ , we see that $a^2 + b^2 = 73$ . There are only two pairs $(a, b)$ with $1 \le a, b \le 9$ satisfying this equality, namely, $(a, b) = (8, 3)$ or $(3, 8)$ . We then get $P = xy = 83 \cdot 38 = 3154$ and this gives the desired answer.

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