Competition · AMC preparation · step 4 of 4

AMC 10 · 2024A · #23

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

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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 the two equations

Subtract equation (2) from (1) and factor by grouping: the left side collapses to (a-c)(b-1) = 13.

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

2STEP 2

Add the two equations

Now add equations (1) and (2) and group the same way: the left side becomes (a+c)(b+1) = 187.

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

3STEP 3

List candidates for b

Integers make 13 prime and 187 = 11·17: so b-1 divides 13, giving b ∈ {-12, 0, 2, 14}, while b+1 must divide 187.

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

4STEP 4

Intersect the two lists

Match each b against the second list: b=2 (→3) and b=14 (→15) fail since neither divides 187, leaving b ∈ {-12, 0}.

Surviving candidates: b ∈ {-12, 0}

5STEP 5

Test b equals 0

Test b=0: it forces a=87 and c=100, but then ca+b = 8700, not 60 — so b=0 is rejected.

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

6STEP 6

Test b equals -12

Test b=-12: the derived equations give a-c = -1 and a+c = -17; adding them yields a = -9, c = -8.

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

7STEP 7

Check all three and finish

Verify (-9, -12, -8) in all three originals, then sum the products: ab+bc+ca = 108+96+72 = 276, choice (D).

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.

Subtract the two equations
Add the two equations
List candidates for b
Intersect the two lists
Test b equals 0
Test b equals -12
Check all three and finish

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