Difference between revisions of "1999 AMC 8 Problems/Problem 15"
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<math>\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60</math> | <math>\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60</math> | ||
| − | ==Solution 1== | + | |
| + | ==Solution== | ||
| + | ===Solution 1=== | ||
There are currently <math>5</math> choices for the first letter, <math>3</math> choices for the second letter, and <math>4</math> choices for the third letter, for a total of <math>5 \cdot 3 \cdot 4 = 60</math> license plates. | There are currently <math>5</math> choices for the first letter, <math>3</math> choices for the second letter, and <math>4</math> choices for the third letter, for a total of <math>5 \cdot 3 \cdot 4 = 60</math> license plates. | ||
| Line 24: | Line 26: | ||
You can get at most <math>100</math> license plates total, giving an additional <math>100 - 60 = 40</math> plates, making the answer <math>\boxed {D}</math> | You can get at most <math>100</math> license plates total, giving an additional <math>100 - 60 = 40</math> plates, making the answer <math>\boxed {D}</math> | ||
| − | ==Solution 2== | + | ===Solution 2=== |
Using the same logic as above, the number of combinations of plates is simply the product of the size of each set of letters. | Using the same logic as above, the number of combinations of plates is simply the product of the size of each set of letters. | ||
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As before, the answer is <math>5 \cdot 5 \cdot 4 - 5 \cdot 3 \cdot 4 = 40</math>, and the correct choice is <math>\boxed{D}</math> | As before, the answer is <math>5 \cdot 5 \cdot 4 - 5 \cdot 3 \cdot 4 = 40</math>, and the correct choice is <math>\boxed{D}</math> | ||
| − | ==See | + | ===Solution 3=== |
| + | Before new letters were added, five different letters could have been chosen for the first position, three for the second, and four for the third. This means that <math>5\cdot 3\cdot 4=60</math> plates could have been made. | ||
| + | |||
| + | If two letters are added to the second set, then <math>5\cdot 5\cdot 4=100</math> plates can be made. If one letter is added to each of the second and third sets, then <math>5\cdot 4\cdot 5=100</math> plates can be made. None of the other four ways to place the two letters will create as many plates. So, <math>100-60=\boxed{40}</math> ADDITIONAL plates can be made.So the correct choice is <math>\boxed{D}</math> | ||
| + | |||
| + | ===Video Solution=== | ||
| + | https://youtu.be/t_tvXQ_30ZA Soo, DRMS, NM | ||
| + | |||
| + | ==See Also== | ||
{{AMC8 box|year=1999|num-b=14|num-a=16}} | {{AMC8 box|year=1999|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:50, 21 December 2022
Contents
Problem
Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?
Solution
Solution 1
There are currently 
choices for the first letter, 
choices for the second letter, and 
choices for the third letter, for a total of 
license plates.
Adding 
letters to the start gives 
plates.
Adding letters to the middle gives 
plates.
Adding letters to the end gives 
plates.
Adding a letter to the start and middle gives 
plates.
Adding a letter to the start and end gives 
plates.
Adding a letter to the middle and end gives 
plates.
You can get at most 
license plates total, giving an additional 
plates, making the answer 
Solution 2
Using the same logic as above, the number of combinations of plates is simply the product of the size of each set of letters.
In general, when three numbers have the same fixed sum, their product will be maximal when they are as close together as possible. This is a 3D analogue of the fact that a rectangle with fixed perimeter maximizes its area when the sides are equal (ie when it becomes a square). In this case, no matter where you add the letters, there will be 
letters in total. If you divide them as evenly as possible among the three groups, you get 
, which is a possible situation.
As before, the answer is 
, and the correct choice is 
Solution 3
Before new letters were added, five different letters could have been chosen for the first position, three for the second, and four for the third. This means that 
plates could have been made.
If two letters are added to the second set, then 
plates can be made. If one letter is added to each of the second and third sets, then 
plates can be made. None of the other four ways to place the two letters will create as many plates. So, 
ADDITIONAL plates can be made.So the correct choice is
Video Solution
https://youtu.be/t_tvXQ_30ZA Soo, DRMS, NM
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
