Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1999 AMC 8 Problems/Problem 15"

(→‎Solution 2)
(→‎Solution 3)
(One intermediate revision by one other user not shown)
Line 25: Line 25:
  
 
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===
 +
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 <math>5 + 3 + 4 + 2 = 14</math> letters in total.  If you divide them as evenly as possible among the three groups, you get <math>5, 5, 4</math>, which is a possible situation.
 +
 +
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>
 +
 +
===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>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=1999|num-b=14|num-a=16}}
 
{{AMC8 box|year=1999|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:26, 9 December 2017

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?

$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$

Solution

Solution 1

There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \cdot 3 \cdot 4 = 60$ license plates.

Adding $2$ letters to the start gives $7\cdot 3 \cdot 4 = 84$ plates.

Adding $2$ letters to the middle gives $5 \cdot 5 \cdot 4 = 100$ plates.

Adding $2$ letters to the end gives $5 \cdot 3 \cdot 6 = 90$ plates.

Adding a letter to the start and middle gives $6 \cdot 4 \cdot 4 = 96$ plates.

Adding a letter to the start and end gives $6 \cdot 3 \cdot 5 = 90$ plates.

Adding a letter to the middle and end gives $5 \cdot 4 \cdot 5 = 100$ plates.

You can get at most $100$ license plates total, giving an additional $100 - 60 = 40$ plates, making the answer $\boxed {D}$


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 $5 + 3 + 4 + 2 = 14$ letters in total. If you divide them as evenly as possible among the three groups, you get $5, 5, 4$, which is a possible situation.

As before, the answer is $5 \cdot 5 \cdot 4 - 5 \cdot 3 \cdot 4 = 40$, and the correct choice is $\boxed{D}$

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 $5\cdot 3\cdot 4=60$ plates could have been made.

If two letters are added to the second set, then $5\cdot 5\cdot 4=100$ plates can be made. If one letter is added to each of the second and third sets, then $5\cdot 4\cdot 5=100$ plates can be made. None of the other four ways to place the two letters will create as many plates. So, $100-60=\boxed{40}$ ADDITIONAL plates can be made.So the correct choice is $\boxed{D}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_15&oldid=88848"