Base | Representation |
---|---|
bin | 110010001111001000… |
… | …100011011000011001 |
3 | 12011020020111021122111 |
4 | 302033020203120121 |
5 | 1340432343212023 |
6 | 40440302352321 |
7 | 3616464601633 |
oct | 621710433031 |
9 | 164206437574 |
10 | 53940991513 |
11 | 20970313332 |
12 | a55488b6a1 |
13 | 51183a0876 |
14 | 2879cc3a53 |
15 | 160a86100d |
hex | c8f223619 |
53940991513 has 2 divisors, whose sum is σ = 53940991514. Its totient is φ = 53940991512.
The previous prime is 53940991453. The next prime is 53940991549. The reversal of 53940991513 is 31519904935.
It is a happy number.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 53940991504 + 9 = 232252^2 + 3^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-53940991513 is a prime.
It is not a weakly prime, because it can be changed into another prime (53940991553) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26970495756 + 26970495757.
It is an arithmetic number, because the mean of its divisors is an integer number (26970495757).
Almost surely, 253940991513 is an apocalyptic number.
It is an amenable number.
53940991513 is a deficient number, since it is larger than the sum of its proper divisors (1).
53940991513 is an equidigital number, since it uses as much as digits as its factorization.
53940991513 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 656100, while the sum is 49.
The spelling of 53940991513 in words is "fifty-three billion, nine hundred forty million, nine hundred ninety-one thousand, five hundred thirteen".
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