AMC 10 · 2005 · #24

Grade 8 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

difference-of-squaresdigit-decompositionperfect-squares convert-to-algebracasework ↑ Prerequisites: difference-of-squaresperfect-squaresdigit-decomposition

📏 Long solution 💡 3 insights

Problem

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 $x + y + m$ ?

Pick an answer.

(A)

(B)

112

(C)

116

(D)

144

(E)

154

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A two-digit number x has its digits reversed to make another two-digit number y. Their squares differ by a perfect square m^2. Find the total x + y + m.

Givens: x and y are two-digit integers; y is x with its two digits swapped; x^2 - y^2 = m^2 for some positive integer m

Unknowns: the value of x + y + m

Understand

Restated: A two-digit number x has its digits reversed to make another two-digit number y. Their squares differ by a perfect square m^2. Find the total x + y + m.

Givens: x and y are two-digit integers; y is x with its two digits swapped; x^2 - y^2 = m^2 for some positive integer m

Plan

Primary tool: #4 Introduce a Variable

Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities

Name the two digits so the reversed number becomes an expression. Turning the digit-swap into algebra reveals a clean factorization, and then divisibility rules narrow the digits down to a single possibility.

Execute — Answer: E

#4 Introduce a Variable 6.EE.B.6 Step 1

Name the digits

Let the tens digit be a and the units digit be b.
Then x = 10a + b, and reversing gives y = 10b + a.
Both a and b are digits from 1 to 9 since both x and y are two-digit numbers.

$$x = 10a + b, \quad y = 10b + a$$

💡 A digit-swap is easy to handle once each digit has its own letter.

#13 Convert to Algebra 7.EE.A.2 Step 2

Factor the difference of squares

Compute x^2 - y^2 as (x - y)(x + y).
The difference x - y = 9a - 9b = 9(a - b), and the sum x + y = 11a + 11b = 11(a + b).
Multiplying gives 99(a - b)(a + b).

$$x^2 - y^2 = (x-y)(x+y) = 9(a-b)\cdot 11(a+b) = 99(a-b)(a+b)$$

💡 Rewriting a difference of squares as two factors exposes the hidden 9 and 11.

#3 Eliminate Possibilities 8.EE.A.2 Step 3

Force a perfect square

Since 99 = 3^2 * 11, the expression is 3^2 * 11 * (a - b)(a + b).
For this to be a perfect square, every prime must appear an even number of times, so 11 must divide (a - b)(a + b).
The difference a - b is at most 8, too small to be a multiple of 11, so 11 must divide the sum: a + b = 11.

$$99(a-b)(a+b) = 3^2\cdot 11\cdot(a-b)(a+b) \implies a+b = 11$$

💡 A perfect square needs each prime factor paired up, so the lone 11 must find a partner.

#3 Eliminate Possibilities 8.EE.C.8 Step 4

Pin down the digits

With a + b = 11, the expression becomes 3^2 * 11^2 * (a - b) = 1089(a - b).
This is a perfect square only when a - b is itself a perfect square.
Because a + b = 11 is odd, a - b must be odd, leaving a - b = 1 or a - b = 9.
Solving a + b = 11 with a - b = 9 gives a = 10, which is not a digit, so a - b = 1.
Then a = 6 and b = 5.

$$1089(a-b)=m^2;\ a+b=11,\ a-b=1 \implies a=6,\ b=5$$

💡 Only a square times 1089 stays a square, and parity leaves just one workable gap between the digits.

#13 Convert to Algebra 8.EE.A.2 Step 5

Add up x, y, and m

So x = 65 and y = 56.
Then m^2 = 1089(6 - 5) = 1089, and m = 33.
The total is x + y + m = 65 + 56 + 33 = 154.
The answer is (E).

$$x=65,\ y=56,\ m=\sqrt{1089}=33 \implies 65+56+33 = 154$$

💡 Once the digits are fixed, the three numbers just add straight to the total.

[1] #4 6.EE.B.6 Let the tens digit be a and the units digit be b. Then x = 10a + b, and reversin

[2] #13 7.EE.A.2 Compute x^2 - y^2 as (x - y)(x + y). The difference x - y = 9a - 9b = 9(a - b),

[3] #3 8.EE.A.2 Since 99 = 3^2 * 11, the expression is 3^2 * 11 * (a - b)(a + b). For this to be

[4] #3 8.EE.C.8 With a + b = 11, the expression becomes 3^2 * 11^2 * (a - b) = 1089(a - b). This

[5] #13 8.EE.A.2 So x = 65 and y = 56. Then m^2 = 1089(6 - 5) = 1089, and m = 33. The total is x

Review

Reasonableness: Check directly: 65^2 - 56^2 = 4225 - 3136 = 1089 = 33^2, a genuine perfect square, so m = 33 works. The digits 6 and 5 are valid, and 65 + 56 + 33 = 154 matches choice (E).

Alternative: Instead of the parity argument, just scan two-digit numbers whose digits sum to 11 (29, 38, 47, 56, 65, ...) and test x^2 - y^2 for a perfect square; only 65 and 56 give one (1089 = 33^2), reaching the same total 154.

CCSS standards used (min grade 8)

6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the tens and units digits a and b to write x = 10a + b and y = 10b + a.)
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem (Factoring x^2 - y^2 into 99(a - b)(a + b) to reveal the 9 and 11.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Requiring a perfect square via even prime exponents and taking m = sqrt(1089) = 33.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Solving a + b = 11 together with a - b = 1 to get a = 6 and b = 5.)

⭐ Write a two-digit number as 10a + b, and a difference of reversed squares always becomes 99 times (a - b)(a + b) -- then the primes tell you which digits fit.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Name the digits

Factor the difference of squares

Force a perfect square

Pin down the digits

Add up x, y, and m

Review

CCSS standards used (min grade 8)

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