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 "2015 AMC 10A Problems/Problem 20"

m (→‎Problem)
Line 18: Line 18:
  
 
Also, when adding 4 to 102, you get 106, which has fewer factors than 104, 108, 110, and 112.
 
Also, when adding 4 to 102, you get 106, which has fewer factors than 104, 108, 110, and 112.
−
 
−
==Dispute==
 
−
 
−
Unfortunately, the original problem statement had an error: the problem had not mentioned that the original side lengths had to be positive integers. This caused every answer choice to work, with one of the sides being <math>2</math>, and the other, a half-integer.  E.g., for <math>102</math>, the sides of the rectangle would be <math>2</math> and <math>49/2</math>. This problem has been modified correctly on the wiki for practicing uses.
 
−
 
−
The AMC gave everyone 6 points for this question.
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}}
 
{{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:55, 5 February 2015

Problem

A rectangle has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$, where $A$ and $P$ are positive integers. Which of the following numbers cannot equal $A+P$?

$\textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108$

NOTE: As later announced by the MAA, this problem is incorrect, as it is not stated that the side lengths must be positive integers, so $A+P$ can equal any of the answer choices. This problem was thrown out, and all contestants received full credit for it (whether they answered correctly, incorrectly, or left it blank).

Solution

Let the rectangle's length and width be $a$ and $b$. Its area is $ab$ and the perimeter is $2(a + b)$.

Then $A + P = ab + 2a + 2b$. Factoring, this is $(a + 2)(b + 2) - 4$.

Looking at the answer choices, only $102$ cannot be written this way, because then either $a$ or $b$ would be $0$.

So the answer is $\boxed{\textbf{(B) }102}$.

Also, when adding 4 to 102, you get 106, which has fewer factors than 104, 108, 110, and 112.

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

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=2015_AMC_10A_Problems/Problem_20&oldid=67650"