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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| + | [ONLY FOR CERTAIN CHINESE 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> | ||
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Therefore, the equation is equal to <math>720 \cdot 7! - 720 \cdot 7! = \boxed{\textbf{(B) }0}</math> | Therefore, the equation is equal to <math>720 \cdot 7! - 720 \cdot 7! = \boxed{\textbf{(B) }0}</math> | ||
| − | ~Aray10 | + | [ONLY FOR CERTAIN CHINESE TESTPAPERS] |
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| + | <math>0 - 5! = \boxed{\textbf{(A) }-120}</math> | ||
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| + | ~Aray10 (Main Solution) and RULE101 (Modifications for certain China test papers) | ||
==Solution 2== | ==Solution 2== | ||
Latest revision as of 00:53, 16 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.
Contents
Problem
What is 
[ONLY FOR CERTAIN CHINESE TESTPAPERS]
What is 
Solution 1
Therefore, the equation is equal to 
[ONLY FOR CERTAIN CHINESE TESTPAPERS]
~Aray10 (Main Solution) and RULE101 (Modifications for certain China test papers)
Solution 2
Factoring out 
gives ![\[7!(10\cdot9\cdot8-1\cdot6!).\]](http://latex.artofproblemsolving.com/7/9/5/7959487d7c87c80b5d44388840d0a0c2c3a179e2.png)
Since 
, the answer is 
~Tacos_are_yummy_1
Factoring 
also works, it just makes the expression in the parenthesis a little harder to compute.
Solution 3
Note that must be divisible by 
, and is the only option divisible by .
Solution 4
can be split into two parts, 
and 
.
We can break the into 
The 
part makes 
, and the 
part makes 
, which is 
.
We still have the 7!, and we can multiply it by 
. This is clearly equivalent to , so our solution is 
.
Solution 5
, 
, and 
. Of course, if you're fast enough, you can do 
. Therefore, 
.
-pepper2831
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/DIl3rLQQkQQ?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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. 
