Difference between revisions of "2019 AMC 10B Problems/Problem 7"
(Created page with "sub2pewds") |
(→Video Solution (CREATIVE PROBLEM SOLVING!!!)) |
||
(40 intermediate revisions by 19 users not shown) | |||
Line 1: | Line 1: | ||
− | + | {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #7]] and [[2019 AMC 12B Problems|2019 AMC 12B #5]]}} | |
+ | |||
+ | ==Problem== | ||
+ | |||
+ | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either <math>12</math> pieces of red candy, <math>14</math> pieces of green candy, <math>15</math> pieces of blue candy, or <math>n</math> pieces of purple candy. A piece of purple candy costs <math>20</math> cents. What is the smallest possible value of <math>n</math>? | ||
+ | |||
+ | <math>\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~IronicNinja | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==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\cdot3=21</math>, so the answer has to be <math>\boxed{\textbf{(B) } 21}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/szqeHGv9l7E | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/7xf_g3YQk00 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/U8LzBqzpQaU | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/HISL2-N5NVg?t=2562 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2019|ab=B|num-b=6|num-a=8}} | ||
+ | {{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 10:25, 24 June 2023
- The following problem is from both the 2019 AMC 10B #7 and 2019 AMC 12B #5, so both problems redirect to this page.
Contents
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
~Education, the Study of Everything
Video Solution
~IceMatrix
Video Solution
~savannahsolver
Video Solution
https://youtu.be/HISL2-N5NVg?t=2562
~ pi_is_3.14
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.