2008 AMC 10B Problem 8

P1-10~2 minPrint

Diophantine Equations Basic Counting Use an invariant, parity, or coloring argument

A class collects $\$50$ to buy flowers for a classmate who is in the hospital. Roses cost $\$3$ each, and carnations cost $\$2$ each. No other flowers are to be used. How many different bouquets could be purchased for exactly $\$50$ ?

Show solution without answering

Key insight

Carnations cost an even amount and 50 is even, so the roses must cost an even amount too. The number of roses is 0, 2, 4, up to 16.

Solution

Once you know how many roses there are, the rest of the money must go to carnations. So count the possible numbers of roses.

Carnations cost $\$2$ each, so the money spent on carnations is always even.
The total is $\$50$ , which is even. So the money spent on roses must also be even.
Roses cost $\$3$ each. The cost $3 \times (\text{roses})$ is even only when the number of roses is even.
The roses cannot cost more than $\$50$ . Since $3 \times 16 = 48$ and $3 \times 18 = 54$ , there are at most $16$ roses.
So the number of roses is one of $0, 2, 4, 6, 8, 10, 12, 14, 16$ . That is $9$ choices.
Each one works: the leftover money is even, so it buys a whole number of carnations. For example, $0$ roses and $25$ carnations; $2$ roses ( $\$6$ ) and $22$ carnations ( $\$44$ ); up to $16$ roses ( $\$48$ ) and $1$ carnation.

The answer is $\boxed{\textbf{(C)}\ 9}$ .

Why this works

With two unknown whole numbers and one equation, fix one of them and see what the other must be. Here the even/odd check tells you exactly which numbers of roses leave a whole number of carnations. Then count the list carefully, including both ends ( $0$ roses and $16$ roses).

Concise solution

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

Let $r$ be the number of roses and $c$ the number of carnations, with $r, c \ge 0$ integers and
$3r + 2c = 50.$
Since $50$ and $2c$ are even, $3r$ must be even, so $r$ is even. Also $3r \le 50$ gives $r \le 16$ . Thus $r \in \{0, 2, 4, \dots, 16\}$ , and each such $r$ determines $c = (50 - 3r)/2$ as a nonnegative integer.

That list has $16/2 + 1 = 9$ values, so there are $9$ bouquets.

The answer is $\boxed{\textbf{(C)}\ 9}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 11: Special Factorizations (§11.5)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques