Base | Representation |
---|---|
bin | 11000000110100000010100… |
… | …11010110100101101101001 |
3 | 20221122202110002110201122021 |
4 | 30003100022122310231221 |
5 | 23421323121214031312 |
6 | 304415521355424441 |
7 | 14110062113241532 |
oct | 1403201232645551 |
9 | 227582402421567 |
10 | 53000071236457 |
11 | 1598422755a874 |
12 | 5b3b928755121 |
13 | 2375b67c35417 |
14 | d13307b65689 |
15 | 61d9c0775c07 |
hex | 30340a6b4b69 |
53000071236457 has 4 divisors (see below), whose sum is σ = 53170489472256. Its totient is φ = 52829653000660.
The previous prime is 53000071236449. The next prime is 53000071236473. The reversal of 53000071236457 is 75463217000035.
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 53000071236457 - 23 = 53000071236449 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (53000071236407) 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, 85209117433 + ... + 85209118054.
It is an arithmetic number, because the mean of its divisors is an integer number (13292622368064).
Almost surely, 253000071236457 is an apocalyptic number.
It is an amenable number.
53000071236457 is a deficient number, since it is larger than the sum of its proper divisors (170418235799).
53000071236457 is a wasteful number, since it uses less digits than its factorization.
53000071236457 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 170418235798.
The product of its (nonzero) digits is 529200, while the sum is 43.
The spelling of 53000071236457 in words is "fifty-three trillion, seventy-one million, two hundred thirty-six thousand, four hundred fifty-seven".
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