Base | Representation |
---|---|
bin | 111001010011011001000010… |
… | …0011000110111010110010011 |
3 | 2110002122222112201012000101202 |
4 | 1302212302010120313112103 |
5 | 1012031212320143200201 |
6 | 4544001514445221415 |
7 | 211111601425366304 |
oct | 16246620430672623 |
9 | 2402588481160352 |
10 | 504042403100051 |
11 | 136670201212082 |
12 | 48646964b6186b |
13 | 18832cc137466a |
14 | 8c67ad51927ab |
15 | 3d41476329c6b |
hex | 1ca6c84637593 |
504042403100051 has 2 divisors, whose sum is σ = 504042403100052. Its totient is φ = 504042403100050.
The previous prime is 504042403100023. The next prime is 504042403100057. The reversal of 504042403100051 is 150001304240405.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-504042403100051 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 504042403099987 and 504042403100023.
It is not a weakly prime, because it can be changed into another prime (504042403100057) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 252021201550025 + 252021201550026.
It is an arithmetic number, because the mean of its divisors is an integer number (252021201550026).
Almost surely, 2504042403100051 is an apocalyptic number.
504042403100051 is a deficient number, since it is larger than the sum of its proper divisors (1).
504042403100051 is an equidigital number, since it uses as much as digits as its factorization.
504042403100051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 9600, while the sum is 29.
Adding to 504042403100051 its reverse (150001304240405), we get a palindrome (654043707340456).
The spelling of 504042403100051 in words is "five hundred four trillion, forty-two billion, four hundred three million, one hundred thousand, fifty-one".
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