PUMaC 2012 · 团队赛 · 第 7 题

PUMaC 2012 — Team Round — Problem 7

(3 digits) How many solutions are there to a + b = c , where a, b, c are integers and | a | , | b | , | c | are all less than 2012? – Alan Chang 4

解析

Problem: (3 digits) How many solutions are there to a + b = c , where a, b, c are integers and | a | , | b | , | c | are all less than 2012? Answer: 189 503 2 503 4 503 Solution: Note that the equation can be written as ( a ) + ( b ) = ( c ) . Thanks to Princeton math professor Andrew Wiles, who proved Fermat’s last theorem, we know that this equation has no nontrivial solutions. Thus, for any solution ( a, b, c ), at least one of a, b, c must be zero. √ 1006 2012 2 If a = 0, then b = c , so b = c . Note that 44 < 2012 < 45, so we have the following solutions: 2 S = { (0 , c , c ) | c = 0 , ± 1 , ± 2 , . . . , ± 44 } a 9 √ 4 503 2012 4 If b = 0, then a = c , so a = c . Note that 6 < 2012 < 7, so we have the following solutions: 4 S = { ( c , 0 , c ) | c = 0 , ± 1 , ± 2 , . . . , ± 6 } b 503 1006 2 If c = 0, then a = − b , so a = − b , so we have the following solutions: 2 S = { ( − b , b, 0) | b = 0 , ± 1 , ± 2 , . . . , ± 44 } c We have | S | = | S | = 89 and | S | = 13. (Here, | S | denotes the number of elements in S .) a c b However, note that (0 , 0 , 0) appears in all three solution sets, so: | S ∪ S ∪ S | = | S | + | S | + | S | − 2 = 89 + 13 + 89 − 2 = 189 a b c a b c Author: Alan – Alan Chang 10