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

Revision as of 20:58, 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

Solution 3 (Crazy Rush)

From looking at the problem, it's obvious that $(\pm7,0)$ are already solutions. From squaring and rearranging, we get $n^2-m^2=49.$ We know that the difference between two consecutive squares is always odd, and for each pair of increasing consecutive squares, the difference starts from $3$ and increases by $2$ each time. This means that there is an existing pair, $(\pm n,m)$ of consecutive squares that will satisfy this equation.

Also note that the answer cannot be infinity because the difference between two squares will increase as the two squares get higher, consecutive or not.

Therefore, the solutions $(\pm7,0)$ and $(\pm n,m)$ where $n$ and $m$ are consecutive that have a square difference of $49$ , give the answer of $\boxed{\text{(D) }4}$ ~Tacos_are_yummy_1

Solution 4 (Quick)

Clearly, $(7,0)$ and $(-7,0)$ are solutions. Notice that $49 = 7^2$ . Remembering our Pythagorean triples, we realize $(25, 24)$ is a solution as well, and by extension, so is $(-25, 24)$ . Since 7 is not part of any other Pythagorean triple, we can conclude the answer is $\boxed{\text{(D) }4}$ . ~i_am_suk_at_math_2

@@ Line 56: / Line 56: @@
 Therefore, the solutions <math>(\pm7,0)</math> and <math>(\pm n,m)</math> where <math>n</math> and <math>m</math> are consecutive that have a square difference of <math>49</math>, give the answer of <math>\boxed{\text{(D) }4}</math> ~Tacos_are_yummy_1
+==Solution 4 (Quick)==
+Clearly, <math>(7,0)</math> and <math>(-7,0)</math> are solutions.
+Notice that <math>49 = 7^2</math>. Remembering our Pythagorean triples, we realize <math>(25, 24)</math> is a solution as well, and by extension, so is <math>(-25, 24)</math>. Since 7 is not part of any other Pythagorean triple, we can conclude the answer is  <math>\boxed{\text{(D) }4}</math>.
+~i_am_suk_at_math_2
 ==See also==
 {{AMC10 box|year=2024|ab=A|num-b=10|num-a=12}}
 {{MAA Notice}}

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