Difference between revisions of "2019 AMC 10B Problems/Problem 7"

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Oh honey, when you knocked on my door
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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 <math>n</math> pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of <math>n</math>?
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<math>\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28</math>
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==Solution==
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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 <math>lcm(12,14,15)</math> = 420. Since a piece of purple candy costs 20 cents, the least value of n can be <math>\frac{420}{20} = \boxed{B) 21}</math>
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iron

Revision as of 12:59, 14 February 2019

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

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

iron

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