2019 AMC 10B Problem 7

Every price is a whole number of cents, so the money must split evenly in each case.

1. Let Casper have $M$ cents. He can buy exactly $12$ red candies, so $M$ divides evenly by $12$. In the same way, $M$ divides evenly by $14$ and by $15$.
2. The smallest number that divides evenly by $12$, $14$ and $15$ is their least common multiple. Break each into primes:
$$12 = 2^2 \cdot 3, \qquad 14 = 2 \cdot 7, \qquad 15 = 3 \cdot 5.$$
Take the highest power of each prime: $2^2 \cdot 3 \cdot 5 \cdot 7 = 420$.
3. So $M$ is $420$, or $840$, or $1260$, and so on.
4. Purple candy costs $20$ cents, so $n = M \div 20$. For $M = 420$ this is $n = 21$, a whole number.
5. Larger choices of $M$ give larger $n$ ($840 \div 20 = 42$, and so on), so the smallest is $21$.

Check that $420$ cents really works: red costs $420 \div 12 = 35$ cents, green $420 \div 14 = 30$ cents, blue $420 \div 15 = 28$ cents. All whole numbers.

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

"Exactly enough money to buy $12$ items at a whole-number price" means the money is a multiple of $12$. Several such facts together mean the money is a common multiple, and "smallest possible" points to the least common multiple. Then the purple price turns the money into a number of candies.

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 $\boxed{\textbf{(B)}\ 21}$.