Base | Representation |
---|---|
bin | 11000011011000110… |
… | …00111101010000111 |
3 | 1020211210221101112022 |
4 | 30031203013222013 |
5 | 203323211241403 |
6 | 10005040113355 |
7 | 642634651424 |
oct | 141543075207 |
9 | 36753841468 |
10 | 13112212103 |
11 | 5619558031 |
12 | 265b30585b |
13 | 130c6ccbb3 |
14 | 8c5618d4b |
15 | 51b219d38 |
hex | 30d8c7a87 |
13112212103 has 2 divisors, whose sum is σ = 13112212104. Its totient is φ = 13112212102.
The previous prime is 13112212063. The next prime is 13112212109. The reversal of 13112212103 is 30121221131.
13112212103 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13112212103 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13112212109) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6556106051 + 6556106052.
It is an arithmetic number, because the mean of its divisors is an integer number (6556106052).
Almost surely, 213112212103 is an apocalyptic number.
13112212103 is a deficient number, since it is larger than the sum of its proper divisors (1).
13112212103 is an equidigital number, since it uses as much as digits as its factorization.
13112212103 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 72, while the sum is 17.
Adding to 13112212103 its reverse (30121221131), we get a palindrome (43233433234).
The spelling of 13112212103 in words is "thirteen billion, one hundred twelve million, two hundred twelve thousand, one hundred three".
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