2011 AMC 10A Problem 10

P1-10~3 minPrint

Divisibility & Factors Primes Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed

A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $\$17.71$ . What was the cost of a pencil in cents?

Show solution without answering

Key insight

Factor 1771 cents as 7 × 11 × 23. A majority of 30 is 16 to 30 students, so 23 bought; each bought 7 pencils at 11 cents.

Solution

Work in cents, so the total is $1771$ cents. The total cost is

(number of students who bought) $\times$ (pencils each) $\times$ (cents per pencil) $= 1771$ .

So each of those three numbers is a factor of $1771$ . First find the factors, then use the conditions to decide which is which.

1. Factor 1771. Try small primes. It is odd, so not divisible by $2$ . Its digits add to $16$ , so not by $3$ . It doesn't end in $0$ or $5$ . Try $7$ : $1771 = 7 \times 253$ . Then $253 = 11 \times 23$ . So
$1771 = 7 \times 11 \times 23.$
2. Find the number of students. A majority of $30$ means more than half, so between $16$ and $30$ students. List all the factors of $1771$ : $1, 7, 11, 23, 77, 161, 253, 1771$ . The only one between $16$ and $30$ is $23$ . So $23$ students bought pencils.
3. Split what is left. Then (pencils each) $\times$ (cents per pencil) $= 1771 \div 23 = 77 = 7 \times 11$ .
4. Each student bought more than $1$ pencil, and the price is more than the number of pencils. The only way to write $77$ as a product of two numbers both bigger than $1$ is $7 \times 11$ . The bigger one is the price, so each student bought $7$ pencils at $11$ cents.

Check: $23 \times 7 \times 11 = 1771$ . ✓

The answer is $\boxed{\textbf{(B)}\ 11}$ .

Why this works

When a total is a product of whole numbers you don't know, break the total into primes. Then turn each word condition into a range or an ordering ("majority" means $16$ to $30$ , "greater than" sets which factor is bigger). Those conditions decide which prime goes where.

Concise solution

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

Let $s$ be the number of students who bought pencils, $n$ the number of pencils each bought, and $c$ the cost of a pencil in cents. Then $s \cdot n \cdot c = 1771$ , with $16 \le s \le 30$ (a majority of $30$ ), $n > 1$ , and $c > n$ .

Factor $1771$ : it is divisible by $7$ , giving $1771 = 7 \cdot 253$ , and $253 = 11 \cdot 23$ . So
$1771 = 7 \cdot 11 \cdot 23.$

The only divisor of $1771$ between $16$ and $30$ is $23$ (the others are $1, 7, 11, 77, 161, 253, 1771$ ). Hence $s = 23$ and $n \cdot c = 77 = 7 \cdot 11$ . With $1 < n < c$ , we need $n = 7$ and $c = 11$ .

The answer is $\boxed{\textbf{(B)}\ 11}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Number Theory, Chapter 2: Primes and Composites
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.2, §4.4)