2006 AIME I Problems/Problem 2

Problem

Let set $\mathcal{A}$ be a 90-element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$

Solution

The smallest $S$ is $1+2+ \ldots +90 = 91 \cdot 45 = 4095$ . The largest $S$ is $11+12+ \ldots +100=111\cdot 45=4995$ . All numbers between $4095$ and $4995$ are possible values of S, so the number of possible values of S is $4995-4095+1=901$ .

Alternatively, for ease of calculation, let set $\mathcal{B}$ be a 10-element subset of $\{1,2,3,\ldots,100\}, and let$ T $be the sum of the elements of$ \mathcal{B} $. Note that the number of possible$ S $is the number of possible$ T=5050-S $. The smallest possible$ T $is$ 1+2+ \ldots +10 = 55 $and the largest is$ 91+92+ \ldots + 100 = 955 $, so the number of possible values of T, and therefore S, is$ 955-55+1=\boxed{901}.

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2006 AIME I Problems/Problem 2

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Solution

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