Competition · AMC preparation · step 4 of 4

AMC 10 · 2023A · #23

Grade 8 arithmetic
factorsdivisibility-rulesdigit-sumdifference-of-squares convert-to-algebraidentify-subproblemssystematic-enumeration ↑ Prerequisites: factorsdivisibility-rules
📏 Long solution 💡 3 insights
Problem
A positive integer N has two factor pairs of the form N = ab. In one pair the two factors differ by 20, and in another pair they differ by 23. Find the sum of the digits of N.

Pick an answer.

(A)
9
(B)
13
(C)
15
(D)
17
(E)
19

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

Two factor-pair conditions about a single unknown N scream Tool #13 (Convert to Algebra) — write N = a(a+20) = b(b+23) and equate. Tool #7 (Identify Subproblems) breaks the algebra into a clean chain: (i) match the two products, (ii) rearrange to a difference of squares, (iii) factor the constant on the right side. Tool #2 (Make a Systematic List) finishes by listing the factor pairs of that small constant (129 = 3 · 43) — there are only a couple, so each gives one quick mini-system to solve.

1STEP 1

Translate the clues into algebra

Write each clue as a product: N = a(a+20) = b(b+23), so a² + 20a = b² + 23b.

a² + 20a = b² + 23b
2STEP 2

Complete both squares

Multiply by 4 and complete the square on each side: (2a+20)² - 400 = (2b+23)² - 529.

(2a+20)² - 400 = (2b+23)² - 529
3STEP 3

Isolate the difference of squares

Rearrange into a difference of squares: (2b+23)² - (2a+20)² = 129.

(2b+23)² - (2a+20)² = 129
4STEP 4

Factor the difference of squares

Factor the difference of squares with X = 2b+23, Y = 2a+20: (2b - 2a + 3)(2b + 2a + 43) = 129.

(2b - 2a + 3)(2b + 2a + 43) = 129
5STEP 5

List the factor pairs of 129

List positive factor pairs of 129 = 3·43 (both prime): only 1·129 and 3·43. Since 2b+2a+43 is the larger factor, two systems arise.

129 = 1 · 129 = 3 · 43
6STEP 6

Solve the first system

Solve system (i): a = b+1 turns 2b+2a+43 = 129 into 4b+45 = 129, so b = 21, a = 22, and N = 22·42 = 21·44 = 924.

b = 21, a = 22, N = 22 · 42 = 21 · 44 = 924
7STEP 7

Solve the second system

System (ii) forces a = 0, not a positive divisor, so it is rejected — the unique value is N = 924.

a = 0 rejected; only N = 924 works
8STEP 8

Add the digits

Compute the sum of digits of N = 924: 9 + 2 + 4 = 15, which is choice (C).

9 + 2 + 4 = 15 → (C)
Answer
15
Verify the factor pairs of N = 924 directly. 924 = 21 · 44 (difference 44 - 21 = 23 ✓) and 924 = 22 · 42 (difference 42 - 22 = 20 ✓). Both required gaps appear in the factor list. The factorization 924 = 2² · 3 · 7 · 11 has (2+1)(1+1)(1+1)(1+1) = 24 divisors, so plenty of factor pairs exist — and both required gaps happen to be among them. The digit sum 9 + 2 + 4 = 15 matches exactly one answer choice.
💡Key takeaway

This AMC 10 problem only needs Grade 8 algebra you already know — write the two clues as N = a(a+20) = b(b+23), rearrange to a difference of squares equaling 129, list its tiny factor pairs, and the only positive solution gives N = 924 with digit sum 15.

  • Translate the clues into algebra
  • Complete both squares
  • Isolate the difference of squares
  • Factor the difference of squares
  • List the factor pairs of 129
  • Solve the first system
  • Solve the second system
  • Add the digits

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