2024 AMC 10B Problems/Problem 2

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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