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s48 - a b has remainder 0 or 1 upon division by 9 and 0 or...

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Solution to Problem 48 Congratulations to this week’s winners Ray Kremer, Felice Kelly, Paul Leisher Correct solutions were also received from Brian Laughlin, Daniel Statman, Burkart Venzke, Gilbert Villalpando, Denis Borris, William Webb, Philippe Fondanaiche, Jennifer Schexnaydre, Al Zimmermann. The only other solution(s) with two digit numbers are (20 + 25) 2 = 2025, and a generously interpreted (98 + 01) 2 = 9801. To arrive at the answer one can simply check all possible pairs of two digit numbers -- a daunting task by hand, but a simple task when armed with just about any programmable calculator. Alternatively, one can proceed as follow. We want to solve the equation (a + b) 2 = 100a + b , where a and b are two digit numbers. Reading the equation modulo 9 and modulo 11 gives the two equations (a + b) 2 = a + b (mod 9) and (a + b) 2 = a + b (mod 11), These equations give you that
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Unformatted text preview: a + b has remainder 0 or 1 upon division by 9, and 0 or 1 upon division by 11. This restricts a + b to be one of 45, 55, or 99; note that (a + b) 2 is a four digit number so a + b is at most 100. Squaring each of these gives the solutions above. Solutions with integers of more than two digits can be handled in either of the ways above -- brute force, or a little number theory. A sampling of some of the more interesting oddities sent in: (4 + 3 + 1 + 649) 2 = 431649 R. Kremer (842 + 72 + 4) 2 = 842724 R. Kremer (585886298 + 179545801) 2 = 585886298179545801 D. Borris a = 25*10 n-2 +,- 5*10 (n/2)-1 , b = 25*10 n-2 , n even A. Zimmermann Finally, from A. Zimmermann comes the tasty (9 + 11 + 25) 3 = 91125; and from Jose Saraiva comes the incredibly filling (17147 + 18793 + 19616) 3 = 171471879319616....
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