1963 AMC 12 Problem 13

Question

If

2^a+2^b=3^c+3^d,

the number of integers

a,

b,

c,

d

which can possibly be negative is, at most:

4

3

2

1

0

A reduced denominator on the left can contain only

2,

while one on the right can contain only

3

Conclude both sides must be integers, then examine when a sum of two negative powers can be integral

In lowest terms, the left side can have only a power of

2

in its denominator, while the right side can have only a power of

3.

Equality therefore forces both sides to be integers. If either

c

or

d

were negative,

3^c+3^d

would retain a factor

3

in its denominator. Thus

c,d\geq0.

Likewise, negative

a

or

b

can give an integer only in the exceptional case

a=b=-1,

whose sum is

1.

But

3^c+3^d\geq2.

Hence

a,b\geq0

as well, so none can be negative.

Thus, the correct answer is E.

Problem 13 in other years