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 $10! - 7! \cdot 6!$

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

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 $720 \cdot 7! - 720 \cdot 7! = \boxed{\textbf{(B) }0}$

Solution for certain China test papers:

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

~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 $10! - 10! =$ $\boxed{\text{(B) }0}$ .

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

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.

Page

Toolbox

Search