2019 AMC 10B Problems/Problem 7

The following problem is from both the 2019 AMC 10B #7 and 2019 AMC 12B #5, so both problems redirect to this page.

Problem

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 $n$ ?

$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$

Solution 1

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 least money he can have is $\text{lcm}{(12,14,15)}$ = 420. Since a piece of purple candy costs 20 cents, the least value of n can be $\frac{420}{20} \implies \boxed{\textbf{(B) } 21}$

Solution 2

We simply need to find a value of $20n$ that divides 12, 14, and 15. $20*18$ divides 12 and 15, but not 14. $20*21$ successfully divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy, so the minimum value of $n$ is $\boxed{\textbf{(B) } 21}$ .

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2019 AMC 10B Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

See Also