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 "2005 AMC 10B Problems/Problem 24"

(→‎Solution 3)
(→‎Video Solution)
 
(3 intermediate revisions by 3 users not shown)
Line 34: Line 34:
  
 
== Solution 4 ==  
 
== Solution 4 ==  
−
Continue the same as solution <math>3</math> until we get <math>33</math>. Knowing that <math>33^2 = 1089</math>, we have narrowed down our Pythagorean triples. We know that the <math>2</math> other squares should be larger than <math>33^2</math>, so we can start testing. If we start testing the <math>40</math>s, it is fruitless since the closest to <math>33^2</math> would be <math>33 - 45 - 54</math> which is not a Pythagorean triple. We can start by testing out the <math>50</math>s, and it turns our that <math>33 - 56 - 65</math> is a Pythagorean triple. Therefore, our answer is <math>33+56+65</math> = $\boxed {(C)144}.
+
Continue the same as Solution <math>3</math> until we get <math>33</math>. Knowing that <math>33^2 = 1089</math>, we have narrowed down our Pythagorean triples. We know that the <math>2</math> other squares should be larger than <math>33^2</math>, so we can start testing.  
 +
 
 +
If we start testing the <math>40</math>s, it is fruitless since the closest to <math>33^2</math> would be <math>33 - 45 - 54</math> which is not a Pythagorean triple. We can start by testing out the <math>50</math>s, and it turns our that <math>33 - 56 - 65</math> is a Pythagorean triple. Therefore, our answer is <math>33+56+65</math> = <math>\boxed{\textbf{(E) }154}</math>.
  
 
~Arcticturn
 
~Arcticturn
 +
 +
==Video Solution==
 +
https://youtu.be/ybsryPSpGfM
 +
 +
~savannahsolver
 +
 +
== Video Solution ==
 +
https://www.youtube.com/watch?v=7Y7OX5uVPac  ~David
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:23, 17 September 2023

Problem

Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits of $x$. The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$. What is $x + y + m$?

$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$

Solution 1

Let $x = 10a+b, y = 10b+a$. The given conditions imply $x>y$, which implies $a>b$, and they also imply that both $a$ and $b$ are nonzero.

Then, $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$.

Since this must be a perfect square, all the exponents in its prime factorization must be even. $99$ factorizes into $3^2 \cdot 11$, so $11|(a-b)(a+b)$. However, the maximum value of $a-b$ is $9-1=8$, so $11|a+b$. The maximum value of $a+b$ is $9+8=17$, so $a+b=11$.

Then, we have $33^2(a-b) = m^2$, so $a-b$ is a perfect square, but the only perfect squares that are within our bound on $a-b$ are $1$ and $4$. We know $a+b=11$, and, for $a-b=1$, adding equations to eliminate $b$ gives us $2a=12 \Longrightarrow a=6, b=5$. Testing $a-b=4$ gives us $2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}$, which is impossible, as $a$ and $b$ must be digits. Therefore, $(a,b) = (6,5)$, and $x+y+m=\boxed{\textbf{(E) }154}$.

Solution 2

The first steps are the same as Solution 1. Let $x = 10a+b, y = 10b+a$, where we know that a and b are digits (whole numbers less than $10$).

Like Solution 1, we end up getting $(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$. This is where the solution diverges.

We know that the left side of the equation is a perfect square because $m$ is an integer. If we factor $99$ into its prime factors, we get $3^2\cdot 11$. In order to get a perfect square on the left side, $(a-b)(a+b)$ must make both prime exponents even. Because the a and b are digits, a simple guess would be that $(a+b)$ (the bigger number) equals $11$ while $(a-b)$ is a factor of nine (1 or 9). The correct guesses are $a = 6, b = 5$ causing $x = 65, y = 56,$ and $m = 33$. The sum of the numbers is $\boxed{\textbf{(E) }154}$

Solution 3

Once again, the solution is quite similar as the above solutions. Since $x$ and $y$ are two digit integers, we can write $x = 10a+b, y = 10b+a$ and because $x^2 - y^2 = (x-y)(x+y)$, substituting and factoring, we get $99(a+b)(a-b) = m^2$.

Therefore, $(a+b)(a-b) = \frac{m^2}{99}$ and $\frac{m^2}{99}$ must be an integer. A quick strategy is to find the smallest such integer $m$ such that $\frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 \cdot 11.$

Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 \cdot 11^2$ gives us 1089 as $n$ and $m = \sqrt{1089} = 33.$ We now get the equation $(x-y)(x+y) = 33^2$, which we can also write as $(x-y)(x+y) = 11^2 \cdot 3^2$.

A very simple guess assumes that $x-y=3^2$ and $x+y=11^2$ since $x$ and $y$ are positive. Finally, we come to the conclusion that $x=65$ and $y=56$, so $x+y+m$ $=$ $\boxed{\textbf{(E) }154}$.

Note that all of the solutions used $a+b$ or $a-b$ as part of their solution.

Solution 4

Continue the same as Solution $3$ until we get $33$. Knowing that $33^2 = 1089$, we have narrowed down our Pythagorean triples. We know that the $2$ other squares should be larger than $33^2$, so we can start testing.

If we start testing the $40$s, it is fruitless since the closest to $33^2$ would be $33 - 45 - 54$ which is not a Pythagorean triple. We can start by testing out the $50$s, and it turns our that $33 - 56 - 65$ is a Pythagorean triple. Therefore, our answer is $33+56+65$ = $\boxed{\textbf{(E) }154}$.

~Arcticturn

Video Solution

https://youtu.be/ybsryPSpGfM

~savannahsolver

Video Solution

https://www.youtube.com/watch?v=7Y7OX5uVPac ~David

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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=2005_AMC_10B_Problems/Problem_24&oldid=198333"