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Difference between revisions of "2024 AMC 10A Problems/Problem 11"
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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
 
<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 ==
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== Solution 1 ==
  
 
Note that <math>m</math> is a nonnegative integer.
 
Note that <math>m</math> is a nonnegative integer.

Revision as of 17:26, 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 1

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 $(m,n).$

Remark

From $(n+m,n-m)=(49,1),(7,7),(-7,-7),(-1,-49),$ we get $(m,n)=(24,25),(0,7),(0,-7),(24,-25),$ respectively.

~MRENTHUSIASM

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

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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