Difference between revisions of "2024 AMC 10B Problems/Problem 2"
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==Problem== | ==Problem== | ||
What is <math>10! - 7! \cdot 6!</math> | What is <math>10! - 7! \cdot 6!</math> | ||
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| + | <math>\textbf{(A) } -120 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 600 \qquad\textbf{(E) } 720</math> | ||
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| + | Certain China testpapers: | ||
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| + | What is <math>10! - 7! \cdot 6! - 5!</math> | ||
<math>\textbf{(A) } -120 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 600 \qquad\textbf{(E) } 720</math> | <math>\textbf{(A) } -120 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 600 \qquad\textbf{(E) } 720</math> | ||
Revision as of 04:10, 14 November 2024
- The following problem is from both the 2024 AMC 10B #2 and 2024 AMC 12B #2, so both problems redirect to this page.
Problem
What is 
Certain China testpapers:
What is 
Solution 1
Therefore, the equation is equal to 
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2024 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
