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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][8, 9].
We can represent 2 three-digit numbers of the problem in the form 100m−1100m - 1, 100n−1100n - 1 with some pairs of integers m,nm, n satisfying 2≤m,n≤102 \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+110000mn - 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−n100-m-n. Since 4≤m+n≤204 \le m+n \le 20, we have 80≤100−m−n≤9680 \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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