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Difference between revisions of "2024 AMC 10A Problems/Problem 11"

(→‎Solution 2)
(→‎Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
−
Squaring both sides of the given equation gives <cmath>n^2-49=m^2\rightarrow n^2-m^2=49\rightarrow (n+m)(n-m)=49</cmath>. Splitting <math>49</math> into its factors (keep in mind it doesn't ask for positive integers, so the factors can be double negative, too) gives six cases:
+
Squaring both sides of the given equation gives <cmath>n^2-49=m^2\rightarrow n^2-m^2=49\rightarrow (n+m)(n-m)=49</cmath> Splitting <math>49</math> into its factors (keep in mind it doesn't ask for positive integers, so the factors can be double negative, too) gives six cases:
  
−
<math>1\cdot49</math>
+
<math>(1\cdot49)</math>
  
−
<math>7\cdot7</math>
+
<math>(7\cdot7)</math>
  
−
<math>49\cdot1</math>
+
<math>(49\cdot1)</math>
  
−
<math>-1\cdot -49</math>
+
<math>(-1\cdot -49)</math>
  
−
<math>-7\cdot -7</math>
+
<math>(-7\cdot -7)</math>
 +
 
 +
<math>(-49\cdot -1)</math>.
  
−
<math>-49\cdot -1</math>.
 
 
Note that the square root in the problem doesn't have <math>\pm</math> with it. Therefore, if there are two solutions, <math>(n,m)</math> and <math>(n,-m)</math>, then these together are to be counted as one solution.
 
Note that the square root in the problem doesn't have <math>\pm</math> with it. Therefore, if there are two solutions, <math>(n,m)</math> and <math>(n,-m)</math>, then these together are to be counted as one solution.
 
The solutions expressed as <math>(n,m)</math> are:
 
The solutions expressed as <math>(n,m)</math> are:
 +
 
<math>(25,24)</math>
 
<math>(25,24)</math>
 +
 
<math>(25,-24)</math>
 
<math>(25,-24)</math>
 +
 
<math>(7,0)</math>
 
<math>(7,0)</math>
 +
 
<math>(-7,0)</math>
 
<math>(-7,0)</math>
 +
 
<math>(-25,24)</math>
 
<math>(-25,24)</math>
 +
 
<math>(-25,-24)</math>.
 
<math>(-25,-24)</math>.
 +
 
<math>(25,24)</math> and <math>(25,-24)</math> are to be counted as one, same for <math>(-25,24)</math> and <math>(-25,-24)</math>. Therefore, the solution is <math>\boxed{\text{(D) }4}</math> ~Tacos_are_yummy_1
 
<math>(25,24)</math> and <math>(25,-24)</math> are to be counted as one, same for <math>(-25,24)</math> and <math>(-25,-24)</math>. Therefore, the solution is <math>\boxed{\text{(D) }4}</math> ~Tacos_are_yummy_1

Revision as of 17:24, 8 November 2024

Problem

How many ordered pairs of integers $(m, n)$ satisfy $\sqrt{n^2 - 49} = m$?

$\textbf{(A)}~1\qquad\textbf{(B)}~2\qquad\textbf{(C)}~3\qquad\textbf{(D)}~4\qquad\textbf{(E)}$ Infinitely many

Solution

Note that $m$ is a nonnegative integer.

We square, rearrange, and apply the difference of squares formula to the given equation: \[(n+m)(n-m)=49.\] It is clear that $n+m\geq n-m,$ so $(n+m,n-m)=(49,1),(7,7),(-7,-7),(-1,-49).$ Each ordered pair $(n+m,n-m)$ gives one ordered pair $(m,n),$ so there are $\boxed{\textbf{(D)}~4}$ such ordered pairs $(m,n).$

Problem

How many ordered pairs of integers $(m, n)$ satisfy $\sqrt{n^2 - 49} = m$?

$\textbf{(A)}~1\qquad\textbf{(B)}~2\qquad\textbf{(C)}~3\qquad\textbf{(D)}~4\qquad\textbf{(E)}$ Infinitely many

Solution 2

Squaring both sides of the given equation gives \[n^2-49=m^2\rightarrow n^2-m^2=49\rightarrow (n+m)(n-m)=49\] Splitting $49$ into its factors (keep in mind it doesn't ask for positive integers, so the factors can be double negative, too) gives six cases:

$(1\cdot49)$

$(7\cdot7)$

$(49\cdot1)$

$(-1\cdot -49)$

$(-7\cdot -7)$

$(-49\cdot -1)$.

Note that the square root in the problem doesn't have $\pm$ with it. Therefore, if there are two solutions, $(n,m)$ and $(n,-m)$, then these together are to be counted as one solution. The solutions expressed as $(n,m)$ are:

$(25,24)$

$(25,-24)$

$(7,0)$

$(-7,0)$

$(-25,24)$

$(-25,-24)$.

$(25,24)$ and $(25,-24)$ are to be counted as one, same for $(-25,24)$ and $(-25,-24)$. Therefore, the solution is $\boxed{\text{(D) }4}$ ~Tacos_are_yummy_1

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