AMC 10 · 2024 · #23
Grade 8 algebranumber-theoryProblem
Integers a, b, and c satisfy ab+c=100, bc+a=87, and ca+b=60. What is ab+bc+ca?
Pick an answer.
AMC 10 2024 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.
Toolkit + CCSS Solution
Understand
Restated: Three integers $a$, $b$, $c$ satisfy $ab + c = 100$, $bc + a = 87$, and $ca + b = 60$. Find the value of $ab + bc + ca$.
Givens: $a, b, c \in \mathbb{Z}$; $ab + c = 100$; $bc + a = 87$; $ca + b = 60$; Answer choices: (A) $212$, (B) $247$, (C) $258$, (D) $276$, (E) $284$
Unknowns: The value of $ab + bc + ca$
Understand
Restated: Three integers $a$, $b$, $c$ satisfy $ab + c = 100$, $bc + a = 87$, and $ca + b = 60$. Find the value of $ab + bc + ca$.
Givens: $a, b, c \in \mathbb{Z}$; $ab + c = 100$; $bc + a = 87$; $ca + b = 60$; Answer choices: (A) $212$, (B) $247$, (C) $258$, (D) $276$, (E) $284$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #3 Eliminate Possibilities, #2 Make a Systematic List
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 $187 = 11 \cdot 17$, 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 $2 \times 2$ linear system.
Execute — Answer: D
6.EE.A.3 Step 1 - Subtract equation (2) from equation (1) and factor by grouping.
- The terms regroup as $b(a-c) - (a-c)$, which is $(a-c)(b-1)$.
💡 Grouping common factors to rewrite the difference as a product is the Grade 6 "generate equivalent expressions" move — and a product equal to a prime is much more informative than a sum.
6.EE.A.3 Step 2 - Add equation (1) and equation (2) and factor by grouping the same way.
- The terms regroup as $b(a+c) + (a+c)$, which is $(a+c)(b+1)$.
💡 Same regrouping trick, just with $+$ instead of $-$. Two equations of the form (something)(b $\pm$ 1) = constant pin down $b$ from two sides.
4.OA.B.4 Step 3 - Use the integer constraint to list candidates for $b$.
- From $(a-c)(b-1) = 13$ and the fact that $13$ is prime, $b-1$ must be one of $\pm 1, \pm 13$, so $b \in \{-12, 0, 2, 14\}$.
- From $(a+c)(b+1) = 187 = 11 \cdot 17$, $b+1$ must be a divisor of $187$, i.e.
- one of $\pm 1, \pm 11, \pm 17, \pm 187$.
💡 Listing the factor pairs of a prime and a product of two primes is a Grade 4 factor-pair drill — and the integer rule says $b-1$ and $b+1$ must come from those lists.
4.OA.B.4 Step 4 - Intersect the two constraints.
- Check each candidate $b$ against the second list: $b=-12$ gives $b+1=-11$ (divides $187$, keep); $b=0$ gives $b+1=1$ (keep); $b=2$ gives $b+1=3$ (not a divisor, drop); $b=14$ gives $b+1=15$ (not a divisor, drop).
💡 Two short divisor lists are a tiny logic table — only the values that appear in both lists survive.
6.EE.B.5 Step 5 - Test $b = 0$ in the originals.
- Then $ab+c = c = 100$ forces $c = 100$, and $bc+a = a = 87$ forces $a = 87$.
- Plug into the third equation: $ca + b = 100 \cdot 87 + 0 = 8700 \ne 60$.
- Rejected.
💡 Substitute the candidate into all three equations and see if it survives — Grade 6 "check whether a value satisfies an equation".
8.EE.C.8 Step 6 - Test $b = -12$.
- Substituting into the two derived equations gives $(a-c)(-13) = 13$ so $a - c = -1$, and $(a+c)(-11) = 187$ so $a + c = -17$.
- Solve this $2 \times 2$ system: adding gives $2a = -18$, so $a = -9$, and then $c = -17 - a = -8$.
💡 Add the two linear equations to cancel $c$ — the Grade 8 elimination method on a $2 \times 2$ linear system.
7.NS.A.2 Step 7 - Verify $(a, b, c) = (-9, -12, -8)$ in all three originals, then compute the target expression.
- Pairwise products: $ab = 108$, $bc = 96$, $ca = 72$.
💡 Multiplying pairs of negatives gives positives, and three positive products sum cleanly — Grade 7 rational-number arithmetic.
6.EE.A.3 Subtract equation (2) from equation (1) and factor by grouping. The terms regrou 6.EE.A.3 Add equation (1) and equation (2) and factor by grouping the same way. The terms 4.OA.B.4 Use the integer constraint to list candidates for $b$. From $(a-c)(b-1) = 13$ an 4.OA.B.4 Intersect the two constraints. Check each candidate $b$ against the second list: 6.EE.B.5 Test $b = 0$ in the originals. Then $ab+c = c = 100$ forces $c = 100$, and $bc+a 8.EE.C.8 Test $b = -12$. Substituting into the two derived equations gives $(a-c)(-13) = 7.NS.A.2 Verify $(a, b, c) = (-9, -12, -8)$ in all three originals, then compute the targ Review
Reasonableness: 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.
Alternative: Tool #6 (Guess and Check) on small integer triples: the three equations grow if $|a|, |b|, |c|$ grow, and the right-hand sides $100, 87, 60$ are in the $60$-$100$ range, suggesting $|a|, |b|, |c| \approx 10$. A short scan over $|b| \le 15$ with $b$ negative (since the targets shrink across the cycle $100 \to 87 \to 60$, suggesting two of the variables are negative so that products are positive but sums are pulled down) quickly lands on $(-9, -12, -8)$. Same answer (D).
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring $ab + c - bc - a$ as $(a-c)(b-1)$ and $ab + c + bc + a$ as $(a+c)(b+1)$ — the factor-by-grouping step that converts the system into two product equations.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing the integer divisors of $13$ (prime) and $187 = 11 \cdot 17$ to enumerate the candidate values $b - 1$ and $b + 1$ can take.)6.EE.B.5Understand solving an equation or inequality as a process of finding values from a specified set (Testing each surviving candidate $b \in \{-12, 0\}$ in the original system to see which one satisfies all three equations.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the $2 \times 2$ linear system $a - c = -1$, $a + c = -17$ by elimination to recover $a = -9$ and $c = -8$ once $b = -12$ is fixed.)7.NS.A.2Apply and extend previous understandings of multiplication and division to multiply and divide rational numbers (Computing the pairwise products $ab = 108$, $bc = 96$, $ca = 72$ with negative integers and summing them to $276$.)
⭐ 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 $187 = 11 \cdot 17$, the integer rule leaves only a couple of possible $b$ values, and from there $a$ and $c$ fall out of a simple $2 \times 2$ system.
⭐ 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 $187 = 11 \cdot 17$, the integer rule leaves only a couple of possible $b$ values, and from there $a$ and $c$ fall out of a simple $2 \times 2$ system.
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