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Number Theory Level 2

x ! = ( 7 ! ) ! 7 ! , x = ? \Large x! = \frac { (7!)!}{7!}, \ \ \ \ \ \ \ x = \ ?


The answer is 5039.

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15 solutions

Discussions for this problem are now closed

Drop TheProblem
Apr 3, 2015

Relevant wiki: Factorials

x ! = ( 7 ! ) ! 7 ! 5040 ⋅ x ! = 5040 ! x ! = 5039 ! x = 5039 \begin{aligned} x! &=\frac{(7!)!}{7!}\\ 5040\cdot x!&=5040! \\ x!&=5039! \\ x&=5039\end{aligned}

297 Helpful 6 Interesting 17 Brilliant 10 Confused

Moderator note:

Great! One could generalize this to show that n ! − 1 = ( n ! ) ! n ! n! - 1 = \dfrac{(n!)!}{n!} , where n = 2 , 3 , 4 , … n=2,3,4,\ldots .

But from 5040 * x! = 5040!, how did it become x! = 5039! ? Please explain?

Diana Kathlyn Cabrera - 5 years ago

Hi Diana

5040 ! = 5040 × 5039 × ⋯ × 2 × 1 5040! = 5040 \times 5039 \times \cdots \times 2 \times 1

→ 5040 × x ! = 5040 × 5039 × ⋯ × 2 × 1 \rightarrow 5040 \times x! = 5040 \times 5039 \times \cdots \times 2 \times 1

Cancelling 5040 on both sides, we get

x ! = 5039 × 5038 × ⋯ × 2 × 1 x! = 5039 \times 5038 \times \cdots \times 2 \times 1 , which is nothing but 5039 ! 5039!

It is pretty easy to generalize and see that x ! = x × ( x − 1 ) ! x! = x \times (x-1)!

Mehul Arora - 5 years ago
Micah Wood
Apr 4, 2015

It's just n ! n \dfrac{n!}{n} , where n = 7 ! n = 7!

Obviously, n ! n = ( n − 1 ) ! \dfrac{n!}{n} = (n-1)! , so ( 7 ! ) ! 7 ! = ( 7 ! − 1 ) ! = 5039 ! \dfrac{(7!)!}{7!} = (7!-1)! = 5039!

x = 5039 x = \boxed{5039}

172 Helpful 4 Interesting 9 Brilliant 3 Confused
Pablo Padilla
Apr 5, 2015

63 Helpful 2 Interesting 4 Brilliant 1 Confused

could you please explain how you did the second step!

Yasir Soltani - 5 years, 8 months ago

x! = x * (x-1)!

Mạnh Hoàng - 5 years, 8 months ago
Paulo Carlos
Apr 4, 2015

7 ! = 5040 7! = 5040

( 7 ! ) ! = 5040 ! (7!)! = 5040!

x ! = 5040 ! 5040 x! \ = \frac {5040!}{5040}

Canceling 5040 in the numerator and in the denominator, it remainders x ! = 5039 ! x! = 5039!

x = 5039 x = \boxed {5039}

14 Helpful 1 Interesting 1 Brilliant 1 Confused

10 Helpful 0 Interesting 1 Brilliant 0 Confused
Otto Bretscher
Apr 5, 2015

A one-liner: x ! = ( 7 ! ) ! 7 ! = 5040 ! 5040 = 5039 ! x!=\frac{(7!)!}{7!}=\frac{5040!}{5040}=5039! , so x = 5039 x=5039 .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

same solution

Austin Joseph - 5 years, 5 months ago

Do you know of any solutions which don't require knowing/computing that 7! = 5040?

Oli Hohman - 5 years, 2 months ago
Amos Tan
Jul 15, 2015

x! = (7!)! / 7! ; 7!= 5040 ; X! = 5040! / 5040 ; 5040!= 1(2 (3( 4 (5 ... (5039 (5040)))))))))...)))); 1(2 (3( 4 (5 ... (5039 (5040)))))))))...)))) / 5040 = 5039! ; X= 5039

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Sajjad Ali
Apr 5, 2015

value of 7!=5040 ; value of (7!)!=5040! ; 

                       x!= (7!)!/7!

                            =(5040!)/5040

                             =(5040*5039*5038................*3*2*1)/5040

                              =5039*5038.............3*2*1

                              =5039!
4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

The expansion of 5040 ! = 5040 × 5039 × 5038 × ⋅ ⋅ ⋅ × 3 × 2 × 1 5040! = 5040 \times 5039 \times 5038 \times \cdot \cdot \cdot \times 3 \times 2 \times 1 is not necessary. A simple recurrence relation n ! n = ( n − 1 ) ! \frac {n!}{n} = (n-1)! should suffice.

Ashwin K
Feb 3, 2016

Question: 

x! = (5040!/5040).

Take for instance,

3!/3 = 6/3 = 2 =2!

In general,

n!/n = n-1! 

Hence the answer is 5039!
3 Helpful 0 Interesting 0 Brilliant 0 Confused

X!= 7!x(7!-1)x(7!-2)x(7!-3)x(7!-4)....3x2x1 divided by 7!= (7!-1)! X!=(7!-1)! hence X = 7! - 1 = 5040-1 = 5039

3 Helpful 0 Interesting 0 Brilliant 0 Confused

( 7 ! ) ! 7 ! \frac{(7!)!}{7!} = ( 7 ! ) ( 7 ! − 1 ) ! 7 ! \frac{(7!)(7!-1)!}{7!} =(7!-1)!=5039!. So x=5039

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Amr Abdelnoor
Mar 27, 2016

x ! = ( 7 ! ) ! 7 ! = 7 ! × ( 7 ! − 1 ) ! 7 ! = ( 7 ! − 1 ) ! x = 7 ! − 1 = 5040 − 1 = 5039 x!\quad =\quad \frac { (7!)! }{ 7! } \quad =\quad \frac { 7!\times (7!-1)! }{ 7! } \quad =\quad (7!-1)!\\ x\quad =\quad 7!-1\quad =\quad 5040-1\quad =\quad 5039

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Davy Ker
Jan 6, 2016

x ! = ( 7 ! ) ! 7 ! 7 ! x ! = ( 7 ! ) ! 7 ! x ! = 7 ! ( 7 ! − 1 ) ! x ! = ( 7 ! − 1 ) ! x = 7 ! − 1 = 5039 x!=\frac{(7!)!}{7!}\\ 7!x!=(7!)!\\ 7!x!=7!(7!-1)!\\ x!=(7!-1)!\\ x=7!-1=5039

Here's a bit of fun: (7!)! is approximately 4.53 × 1 0 16473 4.53 \times 10^{16473}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Maher Farag
Jan 4, 2016

(7!)!=5040! , 7!=5040 -------------------------------- X!=5040(5039!)/5040 -------------------------------- X!=5039! ------------------------------------------------ X = 5039 ---------- proved

0 Helpful 0 Interesting 0 Brilliant 0 Confused

( 7 ! ) ! 7 ! = 7 ! ⋅ ( 7 ! − 1 ) ! 7 ! = ( 7 ! − 1 ) ! \frac{(7!)!}{7!}=\frac{7!\cdot (7!-1)!}{7!}=(7!-1)! = > x ! = ( 7 ! − 1 ) ! = > x = 5040 − 1 = 5039 =>x!=(7!-1)!=>x=5040-1=5039

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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