Base | Representation |
---|---|
bin | 110101011111000110… |
… | …000100010111011001 |
3 | 12111020101200100001021 |
4 | 311133012010113121 |
5 | 1420104040013423 |
6 | 42214420122441 |
7 | 4102111645402 |
oct | 653706042731 |
9 | 174211610037 |
10 | 57430001113 |
11 | 223a0813941 |
12 | b169236421 |
13 | 555318c8a4 |
14 | 2acb4209a9 |
15 | 1761ceb95d |
hex | d5f1845d9 |
57430001113 has 4 divisors (see below), whose sum is σ = 57614663808. Its totient is φ = 57245338420.
The previous prime is 57430001093. The next prime is 57430001117. The reversal of 57430001113 is 31110003475.
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 57430001113 - 229 = 56893130201 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (57430001117) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 92330881 + ... + 92331502.
It is an arithmetic number, because the mean of its divisors is an integer number (14403665952).
Almost surely, 257430001113 is an apocalyptic number.
It is an amenable number.
57430001113 is a deficient number, since it is larger than the sum of its proper divisors (184662695).
57430001113 is a wasteful number, since it uses less digits than its factorization.
57430001113 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 184662694.
The product of its (nonzero) digits is 1260, while the sum is 25.
Adding to 57430001113 its reverse (31110003475), we get a palindrome (88540004588).
The spelling of 57430001113 in words is "fifty-seven billion, four hundred thirty million, one thousand, one hundred thirteen".
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