HMMT 十一月 2009 · 冲刺赛 · 第 9 题
HMMT November 2009 — Guts Round — Problem 9
题目详情
- [ 7 ] Daniel wrote all the positive integers from 1 to n inclusive on a piece of paper. After carefulobservation, he realized that the sum of all the digits that he wrote was exactly 10,000. Find n .
2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUNDnd
英文原题
[ 7 ] Daniel wrote all the positive integers from 1 to n inclusive on a piece of paper. After careful
observation, he realized that the sum of all the digits that he wrote was exactly 10,000. Find n .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND
解析
英文解析
- [ 7 ] Daniel wrote all the positive integers from 1 to n inclusive on a piece of paper. After carefulobservation, he realized that the sum of all the digits that he wrote was exactly 10,000. Find n .
∑x
Answer: 799 Let S ( n ) denote the sum of the digits of n , and let f ( x ) = S ( n ). (We may addn =0
n = 0 because S (0) = 0.) Observe that:
( )
9 9 9 9
∑ ∑ ∑ ∑
f (99) = ( a + b ) = 10 b + 10 a = 900
a =0 a =0
b =0 b =0
If a is an integer between 1 and 9 inclusive, then:
100 a +99 100 a +99
∑ ∑
S ( n ) = ( a + S ( n − 100 a )) = 100 a + f (99) = 100 a + 900
n =100 a n =100 a
Summing, we get:
∑af (100 a + 99) = (100 a + 900) = 900( a + 1) + 50 a ( a + 1)
n =0
This formula can be used to find benchmarks. However, it turns out that this formula alone will suffice,
as things turn out rather nicely:
900( a + 1) + 50 a ( a + 1) = 10000
50 a + 950 a + 900 = 100002
50 a + 950 a − 9100 = 02
50( a + 26)( a − 7) = 0
a = 7
Therefore f (799) = 10000, and our answer is 799.
2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUNDnd