Base | Representation |
---|---|
bin | 11000010001011110… |
… | …01111100010110011 |
3 | 1020122012001222220012 |
4 | 30020233033202303 |
5 | 203142031112201 |
6 | 5553034205135 |
7 | 640634646116 |
oct | 141057174263 |
9 | 36565058805 |
10 | 13031504051 |
11 | 5587a42965 |
12 | 26382837ab |
13 | 12c8a82457 |
14 | 8b8a0a57d |
15 | 5140c65bb |
hex | 308bcf8b3 |
13031504051 has 2 divisors, whose sum is σ = 13031504052. Its totient is φ = 13031504050.
The previous prime is 13031503963. The next prime is 13031504053. The reversal of 13031504051 is 15040513031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13031504051 - 226 = 12964395187 is a prime.
Together with 13031504053, 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 (13031504053) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6515752025 + 6515752026.
It is an arithmetic number, because the mean of its divisors is an integer number (6515752026).
Almost surely, 213031504051 is an apocalyptic number.
13031504051 is a deficient number, since it is larger than the sum of its proper divisors (1).
13031504051 is an equidigital number, since it uses as much as digits as its factorization.
13031504051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 900, while the sum is 23.
The spelling of 13031504051 in words is "thirteen billion, thirty-one million, five hundred four thousand, fifty-one".
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