HMMT 十一月 2017 · 冲刺赛 · 第 22 题

HMMT November 2017 — Guts Round — Problem 22

[ 12 ] A sequence of positive integers a , a , . . . , a has the property that for all integers m where 1 2 2017 ∑ ∑ m m 2 3 1 ≤ m ≤ 2017, 3( a ) = a . Compute a . i 1337 i i =1 i =1

解析

[ 12 ] A sequence of positive integers a , a , . . . , a has the property that for all integers m where 1 2 2017 ∑ ∑ m m 2 3 1 ≤ m ≤ 2017, 3( a ) = a . Compute a . i 1337 i i =1 i =1 Proposed by: Serina Hu Answer: 4011 I claim that a = 3 i for all i . We can conjecture that the sequence should just be the positive multiples i of three because the natural numbers satisfy the property that the square of their sum is the sum of 2 3 their cubes, and prove this by induction. At i = 1, we have that 3 a = a , so a = 3. Now assuming i i i this holds for i = m , we see that ( ) ( ) 2 2 m +1 m ∑ ∑ 3 a = 3 a + a i m +1 i i =1 i =1 m m ∑ ∑ 2 3 = 3 a + a + 6 a a m +1 i m +1 i i =1 i =1 ( ) m ∑ m ( m + 1) 2 3 = 3 a + a + 6 a · 3 m +1 m +1 i 2 i =1 m +1 ∑ 3 = a . i i =1 Therefore, 3 2 2 a = 3 a + a (9 m + 9 m ) m +1 m +1 m +1 2 2 0 = a − 3 a − (9 m + 9 m ) m +1 m +1 0 = ( a − (3 m + 3))( a + 3 m ) m +1 m +1 and because the sequence is positive, a = 3 m + 3, which completes the induction. Then a = m +1 1337 1337 · 3 = 4011.