2006 AIME I Problem 2

Below is the professionally curated solution for Problem 2 of the 2006 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AIME I solutions, or check the answer key.

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Concepts:subsetsextremal argumentcounting integers in a range

Difficulty rating: 1890

2.

Let set A\mathcal{A} be a 9090-element subset of {1,2,3,,100},\{1, 2, 3, \ldots, 100\}, and let SS be the sum of the elements of A.\mathcal{A}. Find the number of possible values of S.S.

Solution:

The smallest possible sum is 1+2++90=4095,1 + 2 + \cdots + 90 = 4095, and the largest is 11+12++100=4995.11 + 12 + \cdots + 100 = 4995.

Every integer in between also occurs. Suppose A\mathcal{A} has sum S4994,S \le 4994, and let kk be the smallest element of A\mathcal{A} with k+1A.k + 1 \notin \mathcal{A}. If kk were 100,100, then A\mathcal{A} would be a block of consecutive integers ending at 100,100, namely {11,,100},\{11, \ldots, 100\}, whose sum exceeds 4994.4994. So k100,k \ne 100, and replacing kk by k+1k + 1 produces a 9090-element subset with sum S+1.S + 1.

Hence SS takes every value from 40954095 to 4995,4995, for 49954095+1=9014995 - 4095 + 1 = 901 possible values.

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