1956 AMC 12 Problem 48

Question

If

p

is a positive integer, then

\dfrac{3p+25}{2p-5}

can be a positive integer, if and only if

p

is:

at least

3

equal to

3,

5,

9,

or

35

no more than

35

equal to

35

equal to

3

or

35

Let

q=2p-5;

positivity forces

q

to be a positive odd integer

Rewrite the quotient as

\frac{3+\frac{65}{q}}{2},

so

q

must be a positive divisor of

65

Set

q=2p-5.

A positive quotient requires

q>0,

and

q

is odd. Since

p=\frac{q+5}{2},

\begin{aligned} \frac{3p+25}{2p-5} &=\frac{3q+65}{2q}\\ &=\frac{3+\frac{65}{q}}{2}. \end{aligned}

Thus

q

must be a positive divisor of

65.

The possibilities

q=1,

q=5,

q=13,

and

q=65

give

p=3,

p=5,

p=9,

and

p=35,

respectively, and each works. No other positive integer

p

works.

Therefore the mathematically correct set is

\{3,5,9,35\},

shown in the adjusted choice B.

Problem 48 in other years