Base | Representation |
---|---|
bin | 10110111010011011… |
… | …01111000110010001 |
3 | 1011202022001210101220 |
4 | 23131031233012101 |
5 | 200143113100423 |
6 | 5352351315253 |
7 | 613560216504 |
oct | 133515570621 |
9 | 34668053356 |
10 | 12301300113 |
11 | 5242843933 |
12 | 247381bb29 |
13 | 12106c930b |
14 | 849a4d13b |
15 | 4bee399e3 |
hex | 2dd36f191 |
12301300113 has 32 divisors (see below), whose sum is σ = 17566761600. Its totient is φ = 7629997056.
The previous prime is 12301300097. The next prime is 12301300129. The reversal of 12301300113 is 31100310321.
It is an interprime number because it is at equal distance from previous prime (12301300097) and next prime (12301300129).
It is not a de Polignac number, because 12301300113 - 24 = 12301300097 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 12301300092 and 12301300101.
It is not an unprimeable number, because it can be changed into a prime (12301300513) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 1575423 + ... + 1583211.
It is an arithmetic number, because the mean of its divisors is an integer number (548961300).
Almost surely, 212301300113 is an apocalyptic number.
It is an amenable number.
12301300113 is a deficient number, since it is larger than the sum of its proper divisors (5265461487).
12301300113 is a wasteful number, since it uses less digits than its factorization.
12301300113 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 8161.
The product of its (nonzero) digits is 54, while the sum is 15.
Adding to 12301300113 its reverse (31100310321), we get a palindrome (43401610434).
The spelling of 12301300113 in words is "twelve billion, three hundred one million, three hundred thousand, one hundred thirteen".
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