2005 AMC 12B Problem 19

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

Answer: E

Concepts:digits difference of squares perfect square

Difficulty rating: 1840

Small hint:

Write

x = 10a + b

and

y = 10b + a

; then

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

Big hint:

For

99(a-b)(a+b)

to be a perfect square,

a+b

must be

11

Solution:

Let

x = 10a + b

and

y = 10b + a

with

a \gt b.

Then

\begin{aligned} &x^2 - y^2 = (10a+b)^2 \\ &\quad {}- (10b+a)^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b). \end{aligned}

Since

99 = 9 \cdot 11,

for this to be a perfect square we need

(a+b)(a-b)

to be a multiple of

11.

As

a + b \le 17

and

a - b \le 8,

the only multiple of

11

available is

a + b = 11.

Then

x^2 - y^2 = 9 \cdot 11^2 (a - b),

which is a perfect square exactly when

a - b

is a perfect square. Because

a+b=11

is odd,

a-b

is odd; and because

1 \le a-b \le 8,

its only possible square value is

1.

Hence

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

So

x = 65,

y = 56,

and

m = \sqrt{65^2 - 56^2}

= \sqrt{1089} = 33.

Thus

x + y + m

= 65 + 56 + 33 = 154.

Thus, the correct answer is E.