A long Factorial!

Number Theory Level 4

Find the sum of all positive integers $n$ such that $n!$ has exactly $n$ digits in decimal representation.

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

The answer is 70.

1 solution

Janardhanan Sivaramakrishnan
Jul 3, 2016

There are $4$ such numbers $1,22,23$ and $24$ . Hence, the sum is $\boxed{70}$ .

It is trivial that $1$ is a valid solution. However, for $2 \le n \le 21$ , it may be seen that $\log_{10} (n!) < n-1$ which means that it has less than $n$ digits.

Further, $\log_{10} 25! \approx 25.19$ hence $25!$ has $26$ digits and every subsequent factorial will add at least one digit. Hence for $n>24$ , $n!$ has more than $n$ digits.

I am interest in how you got in that it may be seen

Mr Yovan - 4 years, 11 months ago

If you want to find number of digits in any number just take its log base 10 and find the ceiling function of that number. You will see from 1 to 21 number of digits are are less except for n=1. After 24 the number of digits start exceeding the number which obviously is going to continue forever as n is increasing by 1 but number will increase in more amount.

Kushagra Sahni - 4 years, 11 months ago

how do you find these numbers ???

can you tell me a method ????

Kushal Bose - 4 years, 11 months ago

A long Factorial!

The answer is 70.

1 solution

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