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

Revision as of 15:14, 14 February 2019

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

iron (don't take credit away from other people @wwt7534)

Solution 2

We simply need to find a value of 20*n 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 $\boxed{n = 21}$ .

Solution 3

This problem is equivalent to finding the LCM of 12, 14, 15, and 20 (and then dividing it by 20). It is easy to see that the prime factorization of said LCM must be $7 \cdot 3 \cdot 5 \cdot 2^2$ . We can divide by 20 now, before we ever multiply it out, leaving us with $7 \cdot 3 = 21 = \boxed{B}$ . However, in this case multiplying it out nets us $420$ , which is worth the time it takes all on its own.

- Robin's solution

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.

@@ Line 17: / Line 17: @@
 ==Solution 3==
-This problem is equivalent to finding the LCM of 12, 14, 15, and 20 (and then dividing it by 20). It is easy to see that the prime factorization of said LCM must be <math>7 \cdot 3 \cdot 5 \cdot 2^2</math>. We can divide by 20 now, before we ever multiply it out, leaving us with <math>7 \cdot 3 = 21 = \boxed{D}</math>. However, in this case multiplying it out nets us <math>420</math>, which is worth the time it takes all on its own.
+This problem is equivalent to finding the LCM of 12, 14, 15, and 20 (and then dividing it by 20). It is easy to see that the prime factorization of said LCM must be <math>7 \cdot 3 \cdot 5 \cdot 2^2</math>. We can divide by 20 now, before we ever multiply it out, leaving us with <math>7 \cdot 3 = 21 = \boxed{B}</math>. However, in this case multiplying it out nets us <math>420</math>, which is worth the time it takes all on its own.
 - Robin's solution

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