PUMaC 2009 · 数论（B 组） · 第 8 题

PUMaC 2009 — Number Theory (Division B) — Problem 8

Let s ( m ) denote the sum of the digits of the positive integer m . Find the largest positive integer that has no digits equal to zero and satisfies the equation s ( n ) 2 2 = s ( n ) 1

解析

Let s ( m ) denote the sum of the digits of the positive integer m . Find the largest positive integer that has no digits equal to zero and satisfies the equation s ( n ) 2 2 = s ( n ) k − 1 k Solution. 1111. Suppose n has k digits, that is 10 ≤ n < 10 , then k ≤ s ( n ) by the 2 2 k 2 2 condition. Also, n < 10 , hence n has at most 2 k digits, and so s ( n ) ≤ 18 k . Thus k s ( n ) 2 2 ≤ 2 = s ( n ) ≤ 18 k s ( n ) which implies k ≤ 6, and so 2 ≤ 18 × 6 or also s ( n ) ≤ 6. If s ( n ) = 6 then n is divisible by 2 2 s ( n ) 3 so n and s ( n ) = 2 is divisible by 3, which is impossible. Hence 1 ≤ s ( n ) ≤ 5 and so 2 2 from the equation, the possible values of s ( n ) are 2,4,8,16, or 32. But the remainder of s ( n ) 2 modulo 9 is the same as the remainder of n modulo 9, which can be only 0,1,4 or 7. Hence s ( n ) is either 2 or 4, and the greatest number satisfying the conditions of the problem is 1111. 3