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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 n2 +, 5*10 (n/2)1 , b = 25*10 n2 , 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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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.
 Spring '10
 AlbertoDelgado
 Combinatorics

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