2026 AMC 10A Problem 13

P11-15~3 minPrint

Basic Counting Set up the equation/formula and compute; no special trick needed

Two bicycle clubs, with disjoint memberships, go on a tour together. On the morning of the tour, members greet each other with a handshake. Everyone shakes hands with everyone else exactly once. There were $119$ handshakes between members of the same club and $112$ handshakes between members of different clubs. How many more people came from the larger club than from the smaller club?

Show solution without answering

Key insight

Within-club plus between-club handshakes is all handshakes: 231, so 22 people. Between-club handshakes are the product of the two club sizes.

Solution

The two given numbers split the handshakes into two kinds. Putting them back together tells us how many people there are, and that is the first thing to find.

1. All handshakes. Every handshake is between two people from the same club or between two people from different clubs, never both. So the total number of handshakes is
$119 + 112 = 231.$

Handshakes among $N$ people. Each person shakes $N - 1$ hands. Multiplying, $N(N-1)$ counts every handshake twice (once for each of the two people), so the total is $\frac{N(N-1)}{2}$ . Small check: $4$ people A, B, C, D make the handshakes AB, AC, AD, BC, BD, CD, which is $6$ , and $\frac{4 \cdot 3}{2} = 6$ .

Find $N$ . We need $\frac{N(N-1)}{2} = 231$ , so $N(N-1) = 462$ . That is two numbers in a row whose product is $462$ . Since $21^2 = 441$ is close to $462$ , try $22 \cdot 21 = 462$ . It works, so there are $N = 22$ people.

4. Handshakes between the clubs. Say the clubs have $a$ and $b$ members, so $a + b = 22$ . A handshake between clubs is one person from the first club with one person from the second club. There are $a$ choices for the first person and $b$ for the second, so
$ab = 112.$

Find $a$ and $b$ . We want two numbers that add to $22$ and multiply to $112$ . List the factor pairs of $112$ and their sums:

Pair	$1 \cdot 112$	$2 \cdot 56$	$4 \cdot 28$	$7 \cdot 16$	$8 \cdot 14$
Sum	$113$	$58$	$32$	$23$	$22$

Only $8$ and $14$ add to $22$ . So the clubs have $14$ and $8$ members.

Check with the other number. Same-club handshakes are $\frac{14 \cdot 13}{2} + \frac{8 \cdot 7}{2} = 91 + 28 = 119$ , which matches the problem.

Answer the question. The larger club has $14 - 8 = 6$ more people.

The answer is $\boxed{\textbf{(C)}\ 6}$ .

Why this works

The two kinds of handshakes together make up all handshakes, so their sum gives the total, and the handshake formula turns the total into the number of people. After that, the between-club count is a simple product, and a sum plus a product pins down two numbers. When a problem splits a count into pieces, try adding the pieces back together before attacking either piece alone.

Alternative approach

Once you know $a + b = 22$ and $ab = 112$ , you can get the difference without finding $a$ and $b$ . Squaring a difference and squaring a sum differ by $4ab$ :
$(a - b)^2 = (a + b)^2 - 4ab = 484 - 448 = 36,$
so the difference is $\sqrt{36} = 6$ .

Concise solution

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

All handshakes number $119 + 112 = 231$ , and $\frac{N(N-1)}{2} = 231$ gives $N(N-1) = 462 = 22 \cdot 21$ , so there are $22$ people. If the clubs have $a \ge b$ members, then $a + b = 22$ , and the between-club handshakes give $ab = 112$ . Among the factor pairs of $112$ ( $1 \cdot 112$ , $2 \cdot 56$ , $4 \cdot 28$ , $7 \cdot 16$ , $8 \cdot 14$ ), only $14 + 8 = 22$ , so $a = 14$ and $b = 8$ . Check: $\binom{14}{2} + \binom{8}{2} = 91 + 28 = 119$ . The difference is $14 - 8 = 6$ .

The answer is $\boxed{\textbf{(C)}\ 6}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Counting & Probability, Chapter 4: Committees and Combinations