2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 7 Easy

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? (2019 AMC 10B Problem, Question#7)

  • A.

    18

  • B.

    21

  • C.

    24

  • D.

    25

  • E.

    28

Answer:B

If he has enough money to buy 12 pieces of red candy, 14 pieces of green candy, and 15 pieces of blue candy, then the smallest amount of money he could have is \rm lcm (12, 14, 15)= 420 cents. Since a piece of purpe candy costs 20 cems,the smallest possible value of n is \frac{420}{20}= \text {(B)}21.

We simply need to find a value of 20n that is divisible by 12, 14, and 15. Observe that 20\cdot 18 is divisible by 12 and 15, but not 14. 20\cdot 21 is divisible by 12, 14, and 15, meaning that we have exact change (in this case, 420 cenis) to buy each type of candy, so the minimum value of n is \text {(B)}21.