Difference between revisions of "2025 AMC 8 Problems"

(Problem 10)
(Problem 11)
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==Problem 11==
 
==Problem 11==
 
Find the smallest positive integer <math>k</math> such that <math>(2^{91}+k)</math> is divisible by <math>127</math>.
 
 
<math>\textbf{(A)}\ 122 \qquad \textbf{(B)}\ 123 \qquad \textbf{(C)}\ 124 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 126</math>
 
  
 
==Problem 12==
 
==Problem 12==

Revision as of 12:06, 26 February 2024

Problem 1

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

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

Problem 2

The 2025 AMC 8 is not held yet. Please do not post false problems.

Problem 3

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Problem 4

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Problem 5

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Problem 6

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Problem 7

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Problem 8

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Problem 9

Problem 10

Problem 11

Problem 12

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Problem 13

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Problem 14

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Problem 15

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Problem 16

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Problem 17

The 2025 AMC 8 is not held yet. Please do not post false problems.

Problem 18

The 2025 AMC 8 is not held yet. Please do not post false problems.

Problem 19

The 2025 AMC 8 is not held yet. Please do not post false problems.

Problem 20

The 2025 AMC 8 is not held yet. Please do not post false problems.

Problem 21

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Problem 22

The 2025 AMC 8 is not held yet. Please do not post false problems.

Problem 23

The 2025 AMC 8 is not held yet. Please do not post false problems.

Problem 24

The 2025 AMC 8 is not held yet. Please do not post false problems.

Problem 25

The 2025 AMC 8 is not held yet. Please do not post false problems.