1999 AMC 8 Problem 15

Attempt Problem 15 of the 1999 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R},\{C,H,L,P,R\}, the second from {A,I,O},\{A,I,O\}, and the third from {D,M,N,T}.\{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?
2424
3030
3636
4040
6060
Answer: D
Concepts:multiplication principleoptimizationcasework
Small hint:
The old number of plates is 5⋅3⋅45\cdot3\cdot4
Big hint:
To maximize a product, make the three factors as balanced as possible
Solution:
There are currently 5⋅3⋅4=605 \cdot 3 \cdot 4 = 60 license plates that can be made.
If both letters are added to the first set, then there are 7⋅3⋅4=84 7 \cdot 3 \cdot 4 = 84 possible plates.
If they are both added to the second, there are 5⋅5⋅4=100 5 \cdot 5 \cdot 4 = 100 plates.
If they are added to the third, there are 5⋅3⋅6=90 5 \cdot 3 \cdot 6 = 90 choices.
If one is added to the first set and the other to the second set, there are 6⋅4⋅4=96 6 \cdot 4 \cdot 4 = 96 plates.
If the other is added to the third set, we get 6⋅3⋅5=90 6 \cdot 3 \cdot 5 = 90 possible plates.
Finally, if the letters are added to the second and third sets, there are 5⋅4⋅5=100 5 \cdot 4 \cdot 5 = 100 plates.
We see that 100100 is the greatest number of plates that we can achieve. This is an additional 100−60=40100 - 60 = 40 plates.
Thus, D is the correct answer.
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