HMMT 二月 2006 · TEAM2 赛 · 第 14 题

HMMT February 2006 — TEAM2 Round — Problem 14

[40] Find the prime factorization of 2 2 2 2006 · 2262 − 669 · 3599 + 1593 · 1337 . (No proof is necessary.)

解析

[40] Find the prime factorization of 2 2 2 2006 · 2262 − 669 · 3599 + 1593 · 1337 . (No proof is necessary.) Answer: 2 · 3 · 7 · 13 · 29 · 59 · 61 · 191 Solution: Upon observing that 2262 = 669 + 1593, 3599 = 1593 + 2006 , and 1337 = 2006 − 669, we are inspired to write a = 2006 , b = 669 , c = − 1593. The expression in 2 2 2 question then rewrites as a ( b − c ) + b ( c − a ) + c ( a − b ). But, by experimenting in the general case (e.g. setting a = b ), we find that this polynomial is zero when two of a, b, c are equal. Immediately we see that it factors as ( b − a )( c − b )( a − c ), so the original expression is a way of writing ( − 1337) · ( − 2262) · (3599). Now, 1337 = 7 · 191, 2 2 2262 = 2 · 3 · 13 · 29, and 3599 = 60 − 1 = 59 · 61.