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April 27[edit]

Multiplication problem[edit]

Hi all. Today I had a puzzle (as every day) in physics that I could not figure out. It went like: if ABCD is a 4 digit integer, with units digit D, tens digit C, etc., and 4*ABCD=DCBA, what are the digits A, B, C, D? At first I tried it by place value (4A*10^3, 3B*10^2...) but I quickly realized that would not work. What is the most efficient (preferably analytic) way to solve this? Thanks. PS: I had only a basic calculator, so WA could not have helped me. 72.128.95.0 (talk) 04:12, 27 April 2011 (UTC)[reply]

The relationship between A and D is that A=1 or 2 (any other value has carryover), and 4 * A = D, and 4 * D = A (mod 10). This means A = 2 and D = 8 (brute force on about 4 possibilities). The problem now is 4*(10B + C) + 3 = 10C + B. The same thing holds with B=0,1,2 since we know that we have no carryover into the thousands digit. But B must be odd, since 4*(10B+C)+3 is odd, so B = 1, and C = 7. Invrnc (talk) 04:29, 27 April 2011 (UTC)[reply]

Well, the answer is right, but the analysis is wrong. 4 * A ≤ D ≤ 4 * A + 3, and 4 * D = A (mod 10). The first has the consequence that A is 1 or 2, the second has the consequence that A is even; hence A = 2. Now, the first has the consequence that D = 8 or 9, and the second has the consequence that D = 3 or 8, hence D = 8.

Given that, 4 * B ≤ C ≤ 4 * B + 3, and 4 * C + 3 = B (mod 10). The first has the consequence that B = 0, 1, or 2, and the second has the consequence that B is odd, so B = 1. Now, the first has the consequence that C = 4, 5, 6, or 7, and the second has the consequence that C = 2 or 7, so C = 7.

Have fun. — Arthur Rubin (talk) 04:55, 30 April 2011 (UTC)[reply]