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Japan Junior Mathematical Olympiad First Round

Algebra Difficulty 3.3 AMC 10/12 Prove it Japan

2 positive integers each with 3 digits are given. Suppose the one's digit and the ten's digit are both 9 for both of these numbers. Write down all possible numbers that can appear as the thousand's digit of the product of these 2 numbers.

Solution

$[8, 9]$ .
We can represent 2 three-digit numbers of the problem in the form $100m - 1$ , $100n - 1$ with some pairs of integers $m, n$ satisfying $2 \le m, n \le 10$ . Then, the product of the 2 given numbers can be written $10000mn - 100(m + n) + 1 = (mn - 1)10000 + (100 - m - n)100 + 1$ , and therefore the value of the thousand's digit of the product equals the ten's digit of the number $100-m-n$ . Since $4 \le m+n \le 20$ , we have $80 \le 100-m-n \le 96$ , from which it follows that the values the thousand's digit of the product of the 2 given numbers can take are 8 and 9.

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