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Difference between revisions of "2024 AMC 10A Problems/Problem 11"
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We square, rearrange, and apply the difference of squares formula to the given equation: <cmath>(n+m)(n-m)=49.</cmath>
 
We square, rearrange, and apply the difference of squares formula to the given equation: <cmath>(n+m)(n-m)=49.</cmath>
 
It is clear that <math>n+m\geq n-m,</math> so <math>(n+m,n-m)=(49,1),(7,7),(-7,-7),(-1,-49).</math> Each ordered pair <math>(n+m,n-m)</math> gives one ordered pair <math>(m,n),</math> so there are <math>\boxed{\textbf{(D)}~4}</math> such ordered pairs <math>(m,n).</math>
 
It is clear that <math>n+m\geq n-m,</math> so <math>(n+m,n-m)=(49,1),(7,7),(-7,-7),(-1,-49).</math> Each ordered pair <math>(n+m,n-m)</math> gives one ordered pair <math>(m,n),</math> so there are <math>\boxed{\textbf{(D)}~4}</math> such ordered pairs <math>(m,n).</math>
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== Problem ==
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How many ordered pairs of integers <math>(m, n)</math> satisfy <math>\sqrt{n^2 - 49} = m</math>?
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<math>\textbf{(A)}~1\qquad\textbf{(B)}~2\qquad\textbf{(C)}~3\qquad\textbf{(D)}~4\qquad\textbf{(E)}</math> Infinitely many
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== Solution 2 ==
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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 [b]positive[/b] integers, so the factors can be double negative, too) gives six cases:
 +
<math>1\cdot49</math>
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<math>7\cdot7</math>
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<math>49\cdot1</math>
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<math>-1\cdot -49</math>
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<math>-7\cdot -7</math>
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<math>-49\cdot -1</math>.
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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:
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<math>(25,24)</math>
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<math>(25,-24)</math>
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<math>(7,0)</math>
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<math>(-7,0)</math>
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<math>(-25,24)</math>
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<math>(-25,-24)</math>.
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<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 16:22, 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 [b]positive[/b] 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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