Mistakes give rise to Problems- 6

Number Theory Level 2

( ( n ! ) ! ) ! = n ! \large ((n!)!)!= n!

We know that n ! = n × ( n − 1 ) × ( n − 2 ) × . . . × 2 × 1 n!= n\times (n-1)\times (n-2) \times ... \times 2\times 1 . And by definition, 0 ! = 1 0!=1 .

How many integers satisfy the equation above?


The answer is 3.

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Aditya Raut by Aditya Raut , 23, India

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5 solutions

Sean Ty
Jul 25, 2014

This one is pretty obvious. The only possible values for n n are 0 0 , 1 1 , and 2 2 .

Because starting from n = 3 n=3 , ( n ! ) ! > n ! (n!)!>n! . While n = 0 , 1 , 2 n=0,1,2 yields ( n ! ) ! = n ! (n!)!=n! .

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Rama Devi
May 19, 2015

The value is same for the integers 0,1,2. Therefore the number of integers satisfying this is 3,which is the required answer

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Noel Lo
Jun 12, 2015

n = 0 , 1 , 2 n = 0, 1, 2 . 2 ! = 2 2! = 2 no matter how many times you use the factorial function as the only number involved in the multiplication other than 2 is 1 and multiplying by 1 does NOT affect the value.

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Soumen Bhowmik
Oct 31, 2014

n= 0, 1, 2

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Sourabh Zunke
Jul 27, 2014

3!=6 so not possible, any integer less than 3 then

0 Helpful 0 Interesting 0 Brilliant 0 Confused

What about negative integers? It doesn't work for negative integers, all of which are less than 3. 3.

Caleb Townsend - 6 years, 3 months ago

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