AMC 10 · 2024 · #23

Grade 8 algebranumber-theory

Archetype Small-Domain Constrained Search

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

📏 Long solution 💡 4 insights

Problem

Integers $a$ , $b$ , and $c$ satisfy $ab + c = 100$ , $bc + a = 87$ , and $ca + b = 60$ . What is $ab + bc + ca$ ?

$\textbf{(A) }212 \qquad \textbf{(B) }247 \qquad \textbf{(C) }258 \qquad \textbf{(D) }276 \qquad \textbf{(E) }284 \qquad$

Pick an answer.

(A)

212

(B)

247

(C)

258

(D)

276

(E)

284

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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

#13 Convert to Algebra 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 $(a-c)(b-1)$.

$$(ab+c) - (bc+a) = 100 - 87 \;\Rightarrow\; (a-c)(b-1) = 13$$

💡 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.

#13 Convert to Algebra 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 $(a+c)(b+1)$.

$$(ab+c) + (bc+a) = 100 + 87 \;\Rightarrow\; (a+c)(b+1) = 187$$

💡 Same regrouping trick, just with $+$ instead of $-$. Two equations of the form (something)(b $\pm$ 1) = constant pin down $b$ from two sides.

#2 Make a Systematic List 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 $\pm 1, \pm 11, \pm 17, \pm 187$.

$$b \in \{-12, 0, 2, 14\} \;\text{ and }\; b+1 \in \{\pm 1, \pm 11, \pm 17, \pm 187\}$$

💡 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.

#3 Eliminate Possibilities 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).

Surviving candidates: $b \in \{-12, 0\}$

💡 Two short divisor lists are a tiny logic table — only the values that appear in both lists survive.

#3 Eliminate Possibilities 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.

$$b = 0 \;\Rightarrow\; (a, c) = (87, 100), \; ca + b = 8700 \ne 60$$

💡 Substitute the candidate into all three equations and see if it survives — Grade 6 "check whether a value satisfies an equation".

#13 Convert to Algebra 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 $c = -17 - a = -8$.

$$a - c = -1, \; a + c = -17 \;\Rightarrow\; a = -9, \; c = -8, \; b = -12$$

💡 Add the two linear equations to cancel $c$ — the Grade 8 elimination method on a $2 \times 2$ linear system.

#13 Convert to Algebra 7.NS.A.2 Step 7

Verify $(a, b, c) = (-9, -12, -8)$ in all three originals, then compute the target expression.
Pairwise products: $ab = 108$, $bc = 96$, $ca = 72$.

$ab + c = 108 - 8 = 100$ ✓, $\; bc + a = 96 - 9 = 87$ ✓, $\; ca + b = 72 - 12 = 60$ ✓; $\; ab + bc + ca = 108 + 96 + 72 = 276 \;\Rightarrow\; \textbf{(D)}$

💡 Multiplying pairs of negatives gives positives, and three positive products sum cleanly — Grade 7 rational-number arithmetic.

[1] #13 6.EE.A.3 Subtract equation (2) from equation (1) and factor by grouping. The terms regrou

[2] #13 6.EE.A.3 Add equation (1) and equation (2) and factor by grouping the same way. The terms

[3] #2 4.OA.B.4 Use the integer constraint to list candidates for $b$. From $(a-c)(b-1) = 13$ an

[4] #3 4.OA.B.4 Intersect the two constraints. Check each candidate $b$ against the second list:

[5] #3 6.EE.B.5 Test $b = 0$ in the originals. Then $ab+c = c = 100$ forces $c = 100$, and $bc+a

[6] #13 8.EE.C.8 Test $b = -12$. Substituting into the two derived equations gives $(a-c)(-13) =

[7] #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.3 Apply 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.4 Find 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.5 Understand 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.8 Analyze 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.2 Apply 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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