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2019 AMC 10B Problem 7

2019 AMC 10B Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

GCD & LCM · difficulty 2 of 5 · about 2 min · position P1-10

Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$ ?

A)
$18$
B)
$21$
C)
$24$
D)
$25$
E)
$28$

Answer

Key insight

Casper's money is a common multiple of 12, 14 and 15, so at least lcm = 420 cents, which buys 420/20 = 21 purple candies.

Solution

Let $M$ be Casper's money in cents. Since $M$ buys exactly $12$ red candies at a whole-cent price, $12 \mid M$ . Likewise $14 \mid M$ and $15 \mid M$ . Therefore $M$ is a multiple of
$\operatorname{lcm}(12, 14, 15) = \operatorname{lcm}(2^2 \cdot 3,\ 2 \cdot 7,\ 3 \cdot 5) = 2^2 \cdot 3 \cdot 5 \cdot 7 = 420.$

Also $M = 20n$ . Every multiple of $420$ is a multiple of $20$ , so $M = 420k$ gives $n = 21k$ . The smallest value is $k = 1$ : $M = 420$ cents, red candy at $35$ cents, green at $30$ , blue at $28$ , purple at $20$ , and $n = 21$ .

The answer is $\boxed{\textbf{(B)}\ 21}$ .

Why this works

"Exactly enough to buy $k$ items at a whole-number price" means the total is divisible by $k$ . Several such statements together say the total is a common multiple, and "smallest possible" points to the least common multiple. The purple-candy price then converts the money into a count.

The trap

Multiplying 12, 14 and 15 together or miscomputing the lcm as 210, then getting a non-integer or too large n.

Common mistakes

Multiplying 12, 14 and 15 together or miscomputing the lcm as 210, then getting a non-integer or too large n.
Forgetting to check that $M$ must also be a multiple of $20$ (here it automatically is, but in general one must include $20$ in the lcm).

Techniques

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