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| − | {{duplicate|[[2025 AMC 10A Problems/Problem 2|2025 AMC 10A #2]] and [[2025 AMC 12A Problems/Problem 2|2025 AMC 12A #2]]}}
| + | Please do not post false problems. |
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| − | ==Problem==
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| − | Let <math>m</math> and <math>n</math> be <math>2</math> integers such that <math>m>n</math>. Suppose <math>m+n=20</math>, <math>m^2+n^2=328</math>, find <math>m^2-n^2</math>.
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| − | <math>\textbf{(A)}~280\qquad\textbf{(B)}~292\qquad\textbf{(C)}~300\qquad\textbf{(D)}~320\qquad\textbf{(E)}~340</math>
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| − | ==Solution==
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| − | Since <math>m^2-n^2=(m+n)(m-n)</math>, we only need to find <math>m-n</math>.
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| − | Squaring <math>m+n=20</math> gives <math>m^2+2mn+n^2=400</math>, thus <math>2mn=400-328=72</math> and <math>mn=36</math>.
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| − | Through trial and error, the only possible value of <math>(m,n)</math> is <math>(18,2)</math>, giving <math>m-n=16</math>.
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| − | Hence, <math>m^2-n^2=20*16=\boxed{\textbf{(D) }320}.</math>
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| − | ~Frankensteinquixotecabin
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| − | ==See also==
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| − | {{AMC10 box|year=2025|ab=A|before=First Problem|num-a=2}}
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| − | {{AMC12 box|year=2025|ab=A|before=First Problem|num-a=2}}
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| − | {{MAA Notice}}
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Latest revision as of 11:57, 3 August 2025
Please do not post false problems.
