A puzzle by A. Vasin from the July-August 1993 issue of Quantum:
Two numbers are mirror numbers if each presents the digits of the other in reverse order, such as 123 and 321. Find two mirror numbers whose product is 92,565.
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This solution is by V. Dubrovsky. The size of the product shows that the factors must have three digits each. So let one of them be abc (or 100a + 10b + c) and the other be cba. The product ends in 5, so either a or c must be 5. Say that’s a. The other factor starts with 5, and 92,565 / 500 < 200, so c must be 1. As to b, we can see that the 6 in 92,565 is the last digit of 5b + b, or 6b, so b must be either 1 or 6, and we can test these candidates to learn that it’s 6. The numbers we seek are 165 and 561.
05/17/2024 UPDATE: Reader Robert Filman points out that we can solve this without actually having to multiply the numbers together. The sum of the digits of any number is the remainder of that number divided by 9, and the sum of the digits of 92,565 mod 9 = 0. So once we’ve established that the end digits of the factors we’re seeking are 1 and 5 and that the middle digit is 1 or 6, as above, we can notice that none of the resulting candidates (115, 165, 511, 561) has a digit sum divisible by 9 and hence each factor must be divisible by 3 — which means that the digit sum of each factor must be a multiple of 3. The only possibilities are 165 and 561. (Thanks, Robert.)
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