Base | Representation |
---|---|
bin | 110001010111101001… |
… | …010110100110001111 |
3 | 12001211100210100122122 |
4 | 301113221112212033 |
5 | 1332031034101212 |
6 | 40204045403155 |
7 | 3554432060546 |
oct | 612751264617 |
9 | 161740710578 |
10 | 53010065807 |
11 | 20532889046 |
12 | a334b694bb |
13 | 4cca569a5a |
14 | 27cc3d715d |
15 | 15a3c76272 |
hex | c57a5698f |
53010065807 has 2 divisors, whose sum is σ = 53010065808. Its totient is φ = 53010065806.
The previous prime is 53010065749. The next prime is 53010065809. The reversal of 53010065807 is 70856001035.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 53010065807 - 26 = 53010065743 is a prime.
It is a super-2 number, since 2×530100658072 (a number of 22 digits) contains 22 as substring.
Together with 53010065809, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (53010065809) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26505032903 + 26505032904.
It is an arithmetic number, because the mean of its divisors is an integer number (26505032904).
Almost surely, 253010065807 is an apocalyptic number.
53010065807 is a deficient number, since it is larger than the sum of its proper divisors (1).
53010065807 is an equidigital number, since it uses as much as digits as its factorization.
53010065807 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 25200, while the sum is 35.
The spelling of 53010065807 in words is "fifty-three billion, ten million, sixty-five thousand, eight hundred seven".
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