AMC 10 · 2012 · #22

Grade 8 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
difference-of-squaressequences-arithmeticdivisor-count convert-to-algebrasystematic-enumeration ↑ Prerequisites: sequences-arithmetic
📏 Medium solution 💡 3 insights
Problem
Add up the first m odd numbers, and separately add up the first n even numbers. The odd total is 212 bigger than the even total. Find every whole number n that makes this possible, then add those n-values together.

Pick an answer.

(A)
255
(B)
256
(C)
257
(D)
258
(E)
259

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

The two sums have clean closed forms, so the whole word problem becomes one equation. The trick is to reshape that equation into a difference of two squares; then the leftover number 847 only has a few factor pairs, and each pair points to one value of n. So the path is: turn the sums into formulas, build the equation, factor it, and list the finite possibilities.

1STEP 1

Turn each sum into a formula

Odds stack into squares (1, 4, 9) and evens into 1x2, 2x3, 3x4, so the sums are m squared and n(n+1).

1+3+5+…+(2m-1) = m², 2+4+6+…+2n = n(n+1)
2STEP 2

Write the equation

The odd sum is 212 more, so m squared = n(n+1) + 212, which expands to m squared = n squared + n + 212.

m² = n(n+1) + 212 = n² + n + 212
3STEP 3

Reshape into a difference of squares

Multiply by 4 to complete the square: 4m squared = (2n+1) squared + 847, so (2m) squared minus (2n+1) squared = 847.

4m² = 4n²+4n+848 = (2n+1)² + 847 → (2m)² - (2n+1)² = 847
4STEP 4

Factor and list the divisor pairs

Factoring gives (2m-2n-1)(2m+2n+1) = 847 = 7x11x11, whose only positive pairs are 1x847, 7x121, 11x77.

(2m-2n-1)(2m+2n+1) = 847 = 7 · 11²; pairs: 1 × 847, 7 × 121, 11 × 77
5STEP 5

Solve each pair for n and add

Larger minus smaller is 4n+2, giving n = 211, 28, 16, each with a whole-number m, so the total is 255.

4n+2 = (larger) - (smaller) → n = 211, 28, 16; 211+28+16 = 255
Answer
255
Check the biggest case: n = 211 gives an even sum of 211 x 212 = 44732, and m = 212 gives an odd sum of 212 squared = 44944; the difference is 44944 - 44732 = 212, exactly as required. The cases n = 28 (m = 32) and n = 16 (m = 22) check the same way. Three valid n-values add to 255, matching choice (A); the nearby choices 256 through 259 would each need an extra or shifted solution that the factor pairs of 847 simply do not provide.
💡Key takeaway

Turn 'odds add to a square' and 'evens add to n(n+1)' into one equation, multiply by 4 to make a difference of squares, and the few factor pairs of 847 reveal every possible n.

  • Turn each sum into a formula
  • Write the equation
  • Reshape into a difference of squares
  • Factor and list the divisor pairs
  • Solve each pair for n and add

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