1998 AIME Problem 7

Question

Let

n

be the number of ordered quadruples

(x_1, x_2, x_3, x_4)

of positive odd integers that satisfy

x_1 + x_2 + x_3 + x_4 = 98.

Find

\frac{n}{100}.

Answer

Substitute

x_i = 2y_i - 1

to turn the odd variables into arbitrary positive integers

The new equation is

y_1 + y_2 + y_3 + y_4 = 51;

count its positive solutions with stars and bars

Write

x_i = 2y_i - 1

where each

y_i

is a positive integer. Then

x_1 + x_2 + x_3 + x_4 = 98

becomes

2(y_1 + y_2 + y_3 + y_4) - 4 = 98,

so

y_1 + y_2 + y_3 + y_4 = 51.

By stars and bars, the number of solutions in positive integers is

\binom{50}{3} = 19600.

Therefore

\frac{n}{100} = 196.

Problem 7 in other years