2000 AIME I Problem 10

Question

A sequence of numbers

x_1,

x_2,

x_3,

\ldots,

x_{100}

has the property that, for every integer

k

between

1

and

100,

inclusive, the number

x_k

is

k

less than the sum of the other

99

numbers. Given that

x_{50} = \frac{m}{n},

where

m

and

n

are relatively prime positive integers, find

m + n.

Answer

If

S

is the sum of all

100

numbers, the condition says

x_k = S - x_k - k

Solve for

x_k

in terms of

S

and

k,

then add up all

100

equations to find

S

Let

S = x_1 + x_2 + \cdots + x_{100}.

The condition says

x_k = (S - x_k) - k,

so

x_k = \frac{S - k}{2}

for every

k.

Summing over

k = 1, \ldots, 100,

\begin{aligned} S &= \frac{100S - (1 + 2 + \cdots + 100)}{2} \\ &= \frac{100S - 5050}{2}, \end{aligned}

so

98S = 5050

and

S = \frac{2525}{49}.

Then

\begin{aligned} x_{50} &= \frac{S - 50}{2} \\ &= \frac{2525 - 2450}{98} = \frac{75}{98}, \end{aligned}

which is in lowest terms, so

m + n = 75 + 98 = 173.

Problem 10 in other years