Difference between revisions of "2024 AMC 10B Problems/Problem 2"

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{{duplicate|[[2024 AMC 10B Problems/Problem 2|2024 AMC 10B #2]] and [[2024 AMC 12B Problems/Problem 2|2024 AMC 12B #2]]}}
  
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==Problem==
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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>
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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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==Solution 1==
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<math>10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!</math>
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<math>6! \cdot 7! = 720 \cdot 7!</math>
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Therefore, the equation is equal to <math>720 \cdot 7! - 720 \cdot 7! = \boxed{\textbf{(B) }0}</math>
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[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)
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==Solution 2==
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Factoring out <math>7!</math> gives <cmath>7!(10\cdot9\cdot8-1\cdot6!).</cmath> Since <math>10\cdot9\cdot8=6!=720</math>, the answer is <math>\boxed{\text{(B) }0}</math> ~Tacos_are_yummy_1
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Factoring <math>6!</math> also works, it just makes the expression in the parenthesis a little harder to compute.
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==Solution 3==
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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>.
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==Solution 4==
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<math>10! - 7! \cdot 6!</math> can be split into two parts, <math>10!</math> and <math>7! \cdot 6!</math>.
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We can break the <math>6!</math> into <math>(2 \cdot 4)(3 \cdot 5 \cdot 6)</math>
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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>.
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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>.
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==Solution 5==
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<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>.
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-pepper2831
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==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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https://youtu.be/DIl3rLQQkQQ?feature=shared
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~ Pi Academy
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/DVlOz24jWuo
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~Thesmartgreekmathdude
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=1|num-a=3}}
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{{AMC12 box|year=2024|ab=B|num-b=1|num-a=3}}
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{{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.

Problem

What is $10! - 7! \cdot 6!$

$\textbf{(A) } -120 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 600 \qquad\textbf{(E) } 720$


[ONLY FOR CERTAIN CHINESE TESTPAPERS]

What is $10! - 7! \cdot 6! - 5!$

$\textbf{(A) } -120 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 600 \qquad\textbf{(E) } 720$

Solution 1

$10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!$

$6! \cdot 7! = 720 \cdot 7!$

Therefore, the equation is equal to $720 \cdot 7! - 720 \cdot 7! = \boxed{\textbf{(B) }0}$

[ONLY FOR CERTAIN CHINESE TESTPAPERS]

$0 - 5! = \boxed{\textbf{(A) }-120}$

~Aray10 (Main Solution) and RULE101 (Modifications for certain China test papers)

Solution 2

Factoring out $7!$ gives \[7!(10\cdot9\cdot8-1\cdot6!).\] Since $10\cdot9\cdot8=6!=720$, the answer is $\boxed{\text{(B) }0}$ ~Tacos_are_yummy_1

Factoring $6!$ also works, it just makes the expression in the parenthesis a little harder to compute.

Solution 3

Note that $10! - 7! \cdot 6!$ must be divisible by $7$, and $\boxed{\text{(B) }0}$ is the only option divisible by $7$.

Solution 4

$10! - 7! \cdot 6!$ can be split into two parts, $10!$ and $7! \cdot 6!$. We can break the $6!$ into $(2 \cdot 4)(3 \cdot 5 \cdot 6)$ The $2 \cdot 4$ part makes $8$, and the $3 \cdot 5 \cdot 6$ part makes $90$, which is $9 \cdot 10$. We still have the 7!, and we can multiply it by $8 \cdot 9 \cdot 10$. This is clearly equivalent to $10!$, so our solution is $10! - 10! =$$\boxed{\text{(B) }0}$.

Solution 5

$10! = 3,628,800$, $7! = 5,040$, and $6! = 720$. Of course, if you're fast enough, you can do $5,040 \cdot 720 = 3,628,800$. Therefore, $3,628,800 - 3,628,800 = \boxed{\text{(B) }0}$.

-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

https://youtu.be/DVlOz24jWuo

~Thesmartgreekmathdude

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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