Base | Representation |
---|---|
bin | 11000111101111010… |
… | …01101101101011011 |
3 | 1021121011111020012122 |
4 | 30132331031231123 |
5 | 204422442442212 |
6 | 10054031531455 |
7 | 653112242456 |
oct | 143675155533 |
9 | 37534436178 |
10 | 13404265307 |
11 | 57593a3a14 |
12 | 2721089b8b |
13 | 1358077844 |
14 | 91232049d |
15 | 536ba9272 |
hex | 31ef4db5b |
13404265307 has 2 divisors, whose sum is σ = 13404265308. Its totient is φ = 13404265306.
The previous prime is 13404265291. The next prime is 13404265309. The reversal of 13404265307 is 70356240431.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13404265307 - 24 = 13404265291 is a prime.
Together with 13404265309, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (13404265309) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6702132653 + 6702132654.
It is an arithmetic number, because the mean of its divisors is an integer number (6702132654).
Almost surely, 213404265307 is an apocalyptic number.
13404265307 is a deficient number, since it is larger than the sum of its proper divisors (1).
13404265307 is an equidigital number, since it uses as much as digits as its factorization.
13404265307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 60480, while the sum is 35.
The spelling of 13404265307 in words is "thirteen billion, four hundred four million, two hundred sixty-five thousand, three hundred seven".
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