When Is Factorial Of A Factorial Redundant?
n ! = ( n ! ) !
How many non-negative integers of n satisfy the equation above?
Notation : ! denotes the factorial notation.
The answer is 3.
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5 solutions
How can 2 satisfy???
Over-rated @Mehul Arora : 3
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Lol. Sab gadhe honge :P
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@Aditya Kumar – Lol could be :P
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@Mehul Arora – Meanwhile see this
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@Aditya Kumar – W0W. Thanks for the hypothetical valentine :P
0! =1 1! =1 2! =2 hence only 3 digit
Is there any way to do this algebraically without trial and error?
a ( a − 1 ) ( a − 2 ) … ( a − n ) = b ≡ b ( b − 1 ) ( b − 2 ) … ( b − n ) = a ∴ a ≡ b If b = a ! then b ! = ( a ! ) ! only through trial and improvement (or by using common sense, i.e. that only small numbers might work) will you know that, by using small natural numbers of a , b ! = ( a ! ) ! . These are 0, 1 and 2 as 0 ! is 1 by convention, 1 × 1 = ( 1 × 1 ) ! and 2 × 1 = ( 2 × 1 ) !
n ! = ( n ! ) !
n = n !
The only numbers which follow the rule directly above are 1 and 2 . 0 follows the initial expression as well since 0 ! = 1 . There are three possible integers hence the answer is 3 .
0 ! = ( 0 ! ) ! → 1 = 1 ! → 1 = 1
1 ! = ( 1 ! ) ! → 1 = 1 ! → 1 = 1
2 ! = ( 2 ! ) ! → 2 = 2 ! → 2 = 2
Honestly, I didn't bother with 2 thinking it was wrong and went with 0, 1 and infinite
Are you certain factorial is defined for Infinity?
We know if a = b , a ! = b ! . So if n ! = ( n ! ) ! , n ! = n . We can easily check few small numbers and get n = 0 , 1 , 2
Um... there is one exception, 0 < 1, but 0! = 1!
n = 0 , 1 , 2 satisfy the given equation