A number theory problem by Anandmay Patel

Number Theory Level 1

( n ! ) 2 = 2 n ! \large (n!)^2=2^{n!}

Find the smallest positive integer n n satisfying the equation above.

Notation : ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 2.

Anandmay Patel by Anandmay Patel , 20, India

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4 solutions

Anand Chitrao
Aug 8, 2016

There is only one positive solution to the problem. If n>2 then n! and hence (n!)^2 contains a power of 3 which is not in 2^(n!). So if such n exists, then it must be 1 or 2. By inspection, the solution turns out to be 2

3 Helpful 0 Interesting 1 Brilliant 0 Confused
Jose Melendrez
Aug 22, 2016

By taking the natural log of both sides : Ln(n!)^2=ln(2)^n! By laws of logarithms , we can move the exponents to the front : 2ln(n!)=n!ln(2) , next notice that the question asks for the smallest possible integer , and by plugging in 1, it makes the left side of the equation zero and therefore not satisfying the equation ,( zero works similarly )we are left with 2 as our answer .

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Niloy Debnath
Sep 2, 2016

First of all, I tried for n=1,simply the both sides can't be equal.Then, I tried for n=2 and got the answer.BTW, it's more easier.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Ku John
Aug 8, 2016

lol guess and check. (0!)^2=1 2^0!=0 (1!)^2=1 2^1!=2 (2!)^2=4 2^2!=4 note:0!=1,1!=1,2!=2(1 2),3!=6(1 2*3)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

2^0! Is not 0 but 2.

Peter van der Linden - 4 years, 10 months ago

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