Base | Representation |
---|---|
bin | 110001011100000011… |
… | …010110111110111001 |
3 | 12002000111220012022112 |
4 | 301130003112332321 |
5 | 1332204000220433 |
6 | 40215250102105 |
7 | 3556320326024 |
oct | 613403267671 |
9 | 162014805275 |
10 | 53083991993 |
11 | 2057059096a |
12 | a35587a935 |
13 | 500c981666 |
14 | 27d815c1bb |
15 | 15aa4ca348 |
hex | c5c0d6fb9 |
53083991993 has 4 divisors (see below), whose sum is σ = 53128714320. Its totient is φ = 53039269668.
The previous prime is 53083991977. The next prime is 53083992037. The reversal of 53083991993 is 39919938035.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 53083991993 - 24 = 53083991977 is a prime.
It is a super-2 number, since 2×530839919932 (a number of 22 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (53083991693) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 22359383 + ... + 22361756.
It is an arithmetic number, because the mean of its divisors is an integer number (13282178580).
Almost surely, 253083991993 is an apocalyptic number.
It is an amenable number.
53083991993 is a deficient number, since it is larger than the sum of its proper divisors (44722327).
53083991993 is a wasteful number, since it uses less digits than its factorization.
53083991993 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 44722326.
The product of its (nonzero) digits is 7085880, while the sum is 59.
The spelling of 53083991993 in words is "fifty-three billion, eighty-three million, nine hundred ninety-one thousand, nine hundred ninety-three".
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