Difference between revisions of "2024 AMC 10B Problems/Problem 2"
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What is <math>10! - 7! \cdot 6! - 5!</math> | What is <math>10! - 7! \cdot 6! - 5!</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> | ||
− | + | [ONLY FOR CERTAIN CHINESE TESTPAPERS] | |
<math>0 - 5! = \boxed{\textbf{(A) }-120}</math> | <math>0 - 5! = \boxed{\textbf{(A) }-120}</math> | ||
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Factoring <math>6!</math> also works, it just makes the expression in the parenthesis a little harder to compute. | Factoring <math>6!</math> also works, it just makes the expression in the parenthesis a little harder to compute. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Note that <math>10! - 7! \cdot 6!</math> must be divisible by <math>7</math>, and <math>\boxed{\text{(B) }0}</math> is the only option divisible by <math>7</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | <math>10! - 7! \cdot 6!</math> can be split into two parts, <math>10!</math> and <math>7! \cdot 6!</math>. | ||
+ | We can break the <math>6!</math> into <math>(2 \cdot 4)(3 \cdot 5 \cdot 6)</math> | ||
+ | The <math>2 \cdot 4</math> part makes <math>8</math>, and the <math>3 \cdot 5 \cdot 6</math> part makes <math>90</math>, which is <math>9 \cdot 10</math>. | ||
+ | We still have the 7!, and we can multiply it by <math>8 \cdot 9 \cdot 10</math>. This is clearly equivalent to <math>10!</math>, so our solution is <math>10! - 10! = </math><math>\boxed{\text{(B) }0}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | <math>10! = 3,628,800</math>, <math>7! = 5,040</math>, and <math>6! = 720</math>. Of course, if you're fast enough, you can do <math>5,040 \cdot 720 = 3,628,800</math>. Therefore, <math>3,628,800 - 3,628,800 = \boxed{\text{(B) }0}</math>. | ||
+ | |||
+ | -pepper2831 | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ||
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==Video Solution 2 by SpreadTheMathLove== | ==Video Solution 2 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=24EZaeAThuE | https://www.youtube.com/watch?v=24EZaeAThuE | ||
+ | |||
+ | == Video Solution by Daily Dose of Math == | ||
+ | |||
+ | https://youtu.be/DVlOz24jWuo | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
+ | |||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2024|ab=B|num-b=1|num-a=3}} | ||
{{AMC12 box|year=2024|ab=B|num-b=1|num-a=3}} | {{AMC12 box|year=2024|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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
[hide]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 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.