2001 AIME I Problem 2

Question

A finite set

\mathcal{S}

of distinct real numbers has the following properties: the mean of

\mathcal{S} \cup \{1\}

is

13

less than the mean of

\mathcal{S},

and the mean of

\mathcal{S} \cup \{2001\}

is

27

more than the mean of

\mathcal{S}.

Find the mean of

\mathcal{S}.

Answer

Let

\mathcal{S}

have

n

elements with mean

x,

and translate each condition into an equation

The conditions give

x - 13(n+1) = 1

and

x + 27(n+1) = 2001.

Subtract them.

Let

\mathcal{S}

have

n

elements with mean

x,

so the elements sum to

nx.

The two conditions say

\frac{nx + 1}{n + 1} = x - 13

and

\frac{nx + 2001}{n + 1} = x + 27.

Clearing denominators,

nx + 1 = (n+1)x - 13(n+1)

gives

x - 13(n+1) = 1,

and

nx + 2001

= (n+1)x + 27(n+1)

gives

x + 27(n+1) = 2001.

Subtracting the first equation from the second yields

40(n+1) = 2000,

so

n + 1 = 50.

Then

x = 1 + 13 \cdot 50 = 651.

Problem 2 in other years