Base | Representation |
---|---|
bin | 111100100011010110… |
… | …1101100011101011011 |
3 | 110102122101112212211202 |
4 | 1321012231230131123 |
5 | 4112303030340322 |
6 | 135423141544415 |
7 | 12252253453616 |
oct | 1710655543533 |
9 | 412571485752 |
10 | 130035402587 |
11 | 5016965a672 |
12 | 2125071510b |
13 | c35431b8cc |
14 | 641808317d |
15 | 35b1008292 |
hex | 1e46b6c75b |
130035402587 has 2 divisors, whose sum is σ = 130035402588. Its totient is φ = 130035402586.
The previous prime is 130035402521. The next prime is 130035402613. The reversal of 130035402587 is 785204530031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 130035402587 - 222 = 130031208283 is a prime.
It is a super-3 number, since 3×1300354025873 (a number of 34 digits) contains 333 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 130035402587.
It is not a weakly prime, because it can be changed into another prime (130035402187) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65017701293 + 65017701294.
It is an arithmetic number, because the mean of its divisors is an integer number (65017701294).
Almost surely, 2130035402587 is an apocalyptic number.
130035402587 is a deficient number, since it is larger than the sum of its proper divisors (1).
130035402587 is an equidigital number, since it uses as much as digits as its factorization.
130035402587 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 100800, while the sum is 38.
The spelling of 130035402587 in words is "one hundred thirty billion, thirty-five million, four hundred two thousand, five hundred eighty-seven".
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