Base | Representation |
---|---|
bin | 11000001000000111001100… |
… | …10011111100010101011011 |
3 | 20221212001000002012000110111 |
4 | 30010003212103330111123 |
5 | 23423224223301341424 |
6 | 304501133544225151 |
7 | 14114055536231662 |
oct | 1404034623742533 |
9 | 227761002160414 |
10 | 53055300027739 |
11 | 159a5698798129 |
12 | 5b4a58076b1b7 |
13 | 237b12a05ba89 |
14 | d15c66ab1dd9 |
15 | 6201541b7194 |
hex | 3040e64fc55b |
53055300027739 has 2 divisors, whose sum is σ = 53055300027740. Its totient is φ = 53055300027738.
The previous prime is 53055300027731. The next prime is 53055300027787. The reversal of 53055300027739 is 93772000355035.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 53055300027739 - 23 = 53055300027731 is a prime.
It is a super-2 number, since 2×530553000277392 (a number of 28 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 53055300027692 and 53055300027701.
It is not a weakly prime, because it can be changed into another prime (53055300027731) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26527650013869 + 26527650013870.
It is an arithmetic number, because the mean of its divisors is an integer number (26527650013870).
Almost surely, 253055300027739 is an apocalyptic number.
53055300027739 is a deficient number, since it is larger than the sum of its proper divisors (1).
53055300027739 is an equidigital number, since it uses as much as digits as its factorization.
53055300027739 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2976750, while the sum is 49.
The spelling of 53055300027739 in words is "fifty-three trillion, fifty-five billion, three hundred million, twenty-seven thousand, seven hundred thirty-nine".
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