Difference between revisions of "2019 AMC 10B Problems/Problem 7"
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If he has enough money to buy <math>12</math> pieces of red candy, <math>14</math> pieces of green candy, and <math>15</math> pieces of blue candy, then the smallest amount of money he could have is <math>\text{lcm}{(12,14,15)} = 420</math> cents. Since a piece of purple candy costs <math>20</math> cents, the smallest possible value of <math>n</math> is <math>\frac{420}{20} = \boxed{\textbf{(B) } 21}</math>. | If he has enough money to buy <math>12</math> pieces of red candy, <math>14</math> pieces of green candy, and <math>15</math> pieces of blue candy, then the smallest amount of money he could have is <math>\text{lcm}{(12,14,15)} = 420</math> cents. Since a piece of purple candy costs <math>20</math> cents, the smallest possible value of <math>n</math> is <math>\frac{420}{20} = \boxed{\textbf{(B) } 21}</math>. | ||
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+ | ~IronicNinja | ||
==Solution 2== | ==Solution 2== | ||
We simply need to find a value of <math>20n</math> that is divisible by <math>12</math>, <math>14</math>, and <math>15</math>. Observe that <math>20 \cdot 18</math> is divisible by <math>12</math> and <math>15</math>, but not <math>14</math>. <math>20 \cdot 21</math> is divisible by <math>12</math>, <math>14</math>, and <math>15</math>, meaning that we have exact change (in this case, <math>420</math> cents) to buy each type of candy, so the minimum value of <math>n</math> is <math>\boxed{\textbf{(B) } 21}</math>. | We simply need to find a value of <math>20n</math> that is divisible by <math>12</math>, <math>14</math>, and <math>15</math>. Observe that <math>20 \cdot 18</math> is divisible by <math>12</math> and <math>15</math>, but not <math>14</math>. <math>20 \cdot 21</math> is divisible by <math>12</math>, <math>14</math>, and <math>15</math>, meaning that we have exact change (in this case, <math>420</math> cents) to buy each type of candy, so the minimum value of <math>n</math> is <math>\boxed{\textbf{(B) } 21}</math>. | ||
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+ | ==Solution 3== | ||
+ | We can notice that the number of purple candy times <math>20</math> has to be divisible by <math>7</math>, because of the <math>14</math> green candies, and <math>3</math>, because of the <math>12</math> red candies. <math>7*3=21</math>, so the answer has to be <math>\boxed{\textbf{(B) } 21}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/7xf_g3YQk00 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/U8LzBqzpQaU | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 13:01, 2 July 2020
- 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 pieces of red candy, pieces of green candy, pieces of blue candy, or pieces of purple candy. A piece of purple candy costs cents. What is the smallest possible value of ?
Solution 1
If he has enough money to buy pieces of red candy, pieces of green candy, and pieces of blue candy, then the smallest amount of money he could have is cents. Since a piece of purple candy costs cents, the smallest possible value of is .
~IronicNinja
Solution 2
We simply need to find a value of that is divisible by , , and . Observe that is divisible by and , but not . is divisible by , , and , meaning that we have exact change (in this case, cents) to buy each type of candy, so the minimum value of is .
Solution 3
We can notice that the number of purple candy times has to be divisible by , because of the green candies, and , because of the red candies. , so the answer has to be .
Video Solution
~IceMatrix
~savannahsolver
See Also
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.