2011 AMC 10A Problem 10

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 $\boxed{\textbf{(B)}\ 11}$.

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.

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 $\boxed{\textbf{(B)}\ 11}$.