AMC 10 · 2024 · #17

Grade 8 algebranumber-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

systems-of-equationsprime-factorizationfactorslinear-equations-two-var convert-to-algebracaseworksystematic-enumeration ↑ Prerequisites: systems-of-equationsprime-factorization

📏 Long solution 💡 4 insights

Problem

Three integers a, b, c satisfy ab + c = 100, bc + a = 87, and ca + b = 60. Find the value of ab + bc + ca.

Pick an answer.

(A)

212

(B)

247

(C)

258

(D)

276

(E)

284

How to solve

Strategy Convert to Algebra

Three equations with three unknowns is the textbook trigger for Tool #13 (Convert to Algebra). Brute-forcing for integer triples would be hopeless, but subtracting and adding pairs of equations factors them by grouping into (a-c)(b-1) = 13 and (a+c)(b+1) = 187. Now the integer constraint becomes powerful: 13 is prime and 187 = 11 · 17, so Tool #3 (Eliminate Possibilities) prunes b to a tiny candidate set. Tool #2 (Systematic List) sweeps the four b values and keeps the one that satisfies both new equations. Once b is pinned, a and c drop out of a 2 × 2 linear system.

1STEP 1

Subtract two equations

The difference factors with product 13.

(ab+c) - (bc+a) = 100 - 87 → (a-c)(b-1) = 13

2STEP 2

Add the same two

The sum factors with product 187.

(ab+c) + (bc+a) = 100 + 87 → (a+c)(b+1) = 187

3STEP 3

Narrow the candidates for b

Only a handful of b satisfy both divisor conditions.

b ∈ {-12, 0, 2, 14} and b+1 ∈ {± 1, ± 11, ± 17, ± 187}

4STEP 4

The survivors

What remains is 0 and negative 12.

Surviving candidates: b ∈ {-12, 0}

5STEP 5

Rule out zero

With b zero the third equation breaks.

b = 0 → (a, c) = (87, 100), ca + b = 8700 ≠ 60

6STEP 6

Pin the three numbers

The triple is -9, -12, -8.

a - c = -1, a + c = -17 → a = -9, c = -8, b = -12

7STEP 7

Verify and sum

All three check out and the sum is 276.

ab + c = 108 - 8 = 100 ✓, bc + a = 96 - 9 = 87 ✓, ca + b = 72 - 12 = 60 ✓; ab + bc + ca = 108 + 96 + 72 = 276 → (D)

Answer

276

The candidate triple (-9, -12, -8) satisfies all three original equations exactly, so it is a genuine solution and not an artifact of the manipulations. The target sum 108 + 96 + 72 = 276 matches choice (D). A magnitude sanity check: the three given right-hand sides 100, 87, 60 have product 522,000, and the products ab, bc, ca sit between 72 and 108 — in the right ballpark for numbers whose pairwise products dominate the small additive terms a, b, c. No other surviving candidate for b exists, so the answer is forced.

💡Key takeaway

Three cyclic equations melt down once you subtract and add pairs — the differences and sums factor into (a-c)(b-1) = 13 and (a+c)(b+1) = 187. Because 13 is prime and 187 = 11 · 17, the integer rule leaves only a couple of possible b values, and from there a and c fall out of a simple 2 × 2 system.