HMMT 二月 2010 · TEAM1 赛 · 第 3 题

HMMT February 2010 — TEAM1 Round — Problem 3

[ 15 ] Call a positive integer in base 10 k-good if we can split it into two integers y and z, such that y is all digits on the left and z is all digits on the right, and such that y = k · z . For example, 2010 is 2-good because we can split it into 20 and 10 and 20 = 2 · 10. 20010 is also 2-good, because we can split it into 20 and 010. In addition, it is 20-good, because we can split it into 200 and 10. Show that there exists a 48-good perfect square.

解析

[ 15 ] Call a positive integer in base 10 k-good if we can split it into two integers y and z, such that y is all digits on the left and z is all digits on the right, and such that y = k · z . For example, 2010 is 2-good because we can split it into 20 and 10 and 20 = 2 · 10. 20010 is also 2-good, because we can split it into 20 and 010. In addition, it is 20-good, because we can split it into 200 and 10. Show that there exists a 48-good perfect square. a a Solution: We wish to find integers a, z such that 48 z · 10 + z = z (48 · 10 + 1) a perfect square, a where z < 10 . This would prove that there exists a 48-good perfect square because we are pulling off the last a digits of the number and get two integers 48 z and z . To make z small by keeping the a product a perfect square, we’d like 48 · 10 + 1 to be divisible by some reasonably large square. Take 42 a a = 42 = ϕ (49). By Euler’s theorem, 10 ≡ 1 (mod 49), so 48 · 10 + 1 is a multiple of 49. Then we ( ) 2 a 42 48 · 10 +1 a a 48 · 10 +1 can take z = . (Clearly z < 10 , so we’re fine.) Then we have z (48 · 10 + 1) = . 49 7