2024 AMC 10A Problem 23

P21-25~6 minPrint

Systems of Equations Diophantine Equations Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Split into exhaustive cases and handle each

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

Show solution without answering

Key insight

Subtract two equations and factor: (a − c)(b − 1) = 13. Since 13 is prime there are only four cases, and only b = −12 works.

Solution

Three equations with products are hard to solve head-on. Subtracting two of them is worth trying, because the result might factor.

1. Subtract the second equation from the first.
$(ab + c) - (bc + a) = 100 - 87 = 13.$
Group the terms: $ab - bc$ has a common factor $b$ , and $c - a$ is the rest. So $b(a - c) - (a - c) = 13$ , which is
$(a - c)(b - 1) = 13.$
2. Use that $13$ is prime. Two integers multiply to $13$ only as $1 \cdot 13$ , $13 \cdot 1$ , $(-1)(-13)$ or $(-13)(-1)$ . So $b - 1$ is $1$ , $13$ , $-1$ or $-13$ , meaning $b$ is $2$ , $14$ , $0$ or $-12$ .
3. Test each case in $ab + c = 100$ .

$b$	$a - c$	so $c =$	$ab + c = 100$ becomes	result
$2$	$13$	$a - 13$	$3a - 13 = 100$	$a = 113/3$ , not an integer
$14$	$1$	$a - 1$	$15a - 1 = 100$	$a = 101/15$ , not an integer
$0$	$-13$	$a + 13$	$a + 13 = 100$	$a = 87$ , $c = 100$
$-12$	$-1$	$a + 1$	$-11a + 1 = 100$	$a = -9$ , $c = -8$

4. Check the survivors in the third equation, $ca + b = 60$ .
- $b = 0$ , $a = 87$ , $c = 100$ : $ca + b = 8700$ . Not $60$ , so this fails.
- $b = -12$ , $a = -9$ , $c = -8$ : $ca + b = 72 - 12 = 60$ . It works. (Also $bc + a = 96 - 9 = 87$ .)
5. Compute what is asked. With $a = -9$ , $b = -12$ , $c = -8$ :
$ab + bc + ca = 108 + 96 + 72 = 276.$

The answer is $\boxed{\textbf{(D)}\ 276}$ .

Why this works

When equations look alike, subtracting them often cancels the messy parts and leaves something that factors. Here the difference factored into two integers multiplying to the prime $13$ , which leaves only four cases. Always check a candidate in all the equations: the case $b = 0$ satisfies two of them but not the third.

Alternative approach

Adding all three equations gives
$(ab + bc + ca) + (a + b + c) = 247.$
This alone does not finish the problem, but it is a good check: with $a + b + c = -9 - 12 - 8 = -29$ , the requested sum is $247 + 29 = 276$ . It also shows why $247$ is among the choices: it is what you get if you forget the $a + b + c$ part.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Subtract the second equation from the first:
$ab + c - bc - a = 13 \quad\Longrightarrow\quad (a - c)(b - 1) = 13.$
Since $13$ is prime, $b - 1 \in \{1, 13, -1, -13\}$ , i.e. $b \in \{2, 14, 0, -12\}$ with $a - c = 13, 1, -13, -1$ respectively. Test each in $ab + c = 100$ :

$b = 2$ , $c = a - 13$ : $2a + a - 13 = 100$ , so $3a = 113$ . Not an integer.
$b = 14$ , $c = a - 1$ : $14a + a - 1 = 100$ , so $15a = 101$ . Not an integer.
$b = 0$ , $c = a + 13$ : $c = 100$ , $a = 87$ . Then $ca + b = 8700 \ne 60$ . Rejected.
$b = -12$ , $c = a + 1$ : $-12a + a + 1 = 100$ , so $a = -9$ and $c = -8$ . Check: $bc + a = 96 - 9 = 87$ and $ca + b = 72 - 12 = 60$ . Valid.

So $(a, b, c) = (-9, -12, -8)$ and
$ab + bc + ca = 108 + 96 + 72 = 276.$

The answer is $\boxed{\textbf{(D)}\ 276}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Algebra, Chapter 11: Special Factorizations
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)