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NL628
NL628
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Uh...I don't think this'll pass as an answer but how about

$\dfrac{(2+2+2+2)!}{2} = 20160$

Lol lucky guess.

EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:

The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.

EDIT: Just cuz I'm bored I'm gonna do it for 6 2's and 4 2's.

6 2's: $\bigg(\dfrac{\log_2(2^{2^{2^2}})}{2}\bigg)!$

or

4 2's: $\Big(\dfrac{2^{2^2}}{2}\Big)!$

or

4 2's: $\dfrac{\Big(2\cdot 2\cdot 2\Big)!}{2}$

Uh...I don't think this'll pass as an answer but how about

$\dfrac{(2+2+2+2)!}{2} = 20160$

Lol lucky guess.

EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:

The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.

EDIT: Just cuz I'm bored I'm gonna do it for 6 2's and 4 2's.

6 2's: $\bigg(\dfrac{\log_2(2^{2^{2^2}})}{2}\bigg)!$

or

4 2's: $\Big(\dfrac{2^{2^2}}{2}\Big)!$

or

4 2's: $\dfrac{\Big(2\cdot 2\cdot 2\Big)!}{2}$

Uh...I don't think this'll pass as an answer but how about

$\dfrac{(2+2+2+2)!}{2} = 20160$

Lol lucky guess.

EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:

The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.

added 213 characters in body
Source Link
NL628
NL628
  • 7.5k
  • 3
  • 35
  • 77

Uh...I don't think this'll pass as an answer but how about

$\dfrac{(2+2+2+2)!}{2} = 20160$

Lol lucky guess.

EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:

The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.

EDIT: Just cuz I'm bored I'm gonna do it for 6 2's and 4 2's.

6 2's: $\bigg(\dfrac{\log_2(2^{2^{2^2}})}{2}\bigg)!$

or

4 2's: $\Big(\dfrac{2^{2^2}}{2}\Big)!$

or

4 2's: $\dfrac{\Big(2\cdot 2\cdot 2\Big)!}{2}$

Uh...I don't think this'll pass as an answer but how about

$\dfrac{(2+2+2+2)!}{2} = 20160$

Lol lucky guess.

EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:

The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.

Uh...I don't think this'll pass as an answer but how about

$\dfrac{(2+2+2+2)!}{2} = 20160$

Lol lucky guess.

EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:

The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.

EDIT: Just cuz I'm bored I'm gonna do it for 6 2's and 4 2's.

6 2's: $\bigg(\dfrac{\log_2(2^{2^{2^2}})}{2}\bigg)!$

or

4 2's: $\Big(\dfrac{2^{2^2}}{2}\Big)!$

or

4 2's: $\dfrac{\Big(2\cdot 2\cdot 2\Big)!}{2}$

Source Link
NL628
NL628
  • 7.5k
  • 3
  • 35
  • 77

Uh...I don't think this'll pass as an answer but how about

$\dfrac{(2+2+2+2)!}{2} = 20160$

Lol lucky guess.

EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:

The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.