2025 AMC 10A Problems: Difference between revisions

Revision as of 06:01, 11 July 2025

2025 AMC 10A (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator). Figures are not necessarily drawn to scale. You will have 75 minutes working time to complete the test.
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Find the smallest positive integer $k$ such that $2^{91}+k$ is divisible by $127$ .

$\textbf{(A)}~122\qquad\textbf{(B)}~123\qquad\textbf{(C)}~124\qquad\textbf{(D)}~125\qquad\textbf{(E)}~126$

Let $m$ and $n$ be $2$ integers such that $m>n$ . Suppose $m+n=20$ , $m^2+n^2=328$ , find $m^2-n^2$ .

$\textbf{(A)}~280\qquad\textbf{(B)}~292\qquad\textbf{(C)}~300\qquad\textbf{(D)}~320\qquad\textbf{(E)}~340$

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 == Problem 2 ==
+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>.
+<math>\textbf{(A)}~280\qquad\textbf{(B)}~292\qquad\textbf{(C)}~300\qquad\textbf{(D)}~320\qquad\textbf{(E)}~340</math>
 == Problem 3 ==

2025 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions