2005 AMC 10B Problem 24

Question

Let

x

and

y

be two-digit integers such that

y

is obtained by reversing the digits of

x.

The integers

x

and

y

satisfy

x^2 - y^2 = m^2

for some positive integer

m.

What is

x + y + m?

88

112

116

144

154

Write

x=10a+b

and

y=10b+a,

then factor

x^2-y^2

x^2-y^2=99(a+b)(a-b)

must be a perfect square, which forces

a+b=11

Write

x = 10a + b

and

y = 10b + a

with

a \gt b.

Then

\begin{aligned} m^2 &= x^2 - y^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b). \end{aligned}

Since

99 = 9 \cdot 11,

for

m^2

to be a perfect square we need

(a+b)(a-b)

to be divisible by

11.

As

a + b \le 17,

this forces

a + b = 11,

and then

a - b

must itself be a perfect square.

With

a - b \le 8,

the only workable case is

a - b = 1,

giving

(a, b) = (6, 5).

Then

x = 65,

y = 56,

and

m^2 = 99 \cdot 11 = 33^2,

so

m = 33.

Therefore

x + y + m

= 65 + 56 + 33

= 154.

Thus, E is the correct answer.

Problem 24 in other years