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Difference between revisions of "2024 AIME II Problems"

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==Problem 7==
 
==Problem 7==
  
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Let <math>N</math> be the greatest four-digit integer with the property that whenever one of its digits is changed to <math>1</math>, the resulting number is divisible by <math>7</math>. Let <math>Q</math> and <math>R</math> be the quotient and remainder, respectively, when <math>N</math> is divided by <math>1000</math>. Find <math>Q+R</math>.
  
 
[[2024 AIME II Problems/Problem 7|Solution]]
 
[[2024 AIME II Problems/Problem 7|Solution]]
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==Problem 11==
 
==Problem 11==
  
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Find the number of triples of nonnegative integers <math>(a, b, c)</math> satisfying <math>a + b + c = 300</math> and
  
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<cmath>a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b = 6,000,000.</cmath>
  
 
[[2024 AIME II Problems/Problem 11|Solution]]
 
[[2024 AIME II Problems/Problem 11|Solution]]

Revision as of 20:13, 8 February 2024

2024 AIME II (Answer Key)
Printable version | AoPS Contest Collections • PDF

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, rulers and compasses are permitted. In particular, graph paper, protractors, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

Solution

Problem 2

Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Let $N$ be the greatest four-digit integer with the property that whenever one of its digits is changed to $1$, the resulting number is divisible by $7$. Let $Q$ and $R$ be the quotient and remainder, respectively, when $N$ is divided by $1000$. Find $Q+R$.

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Find the number of triples of nonnegative integers $(a, b, c)$ satisfying $a + b + c = 300$ and

\[a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b = 6,000,000.\]

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
2024 AIME I 		Followed by
2025 AIME I
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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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