Base | Representation |
---|---|
bin | 1110111000001011110… |
… | …01101010111000000101 |
3 | 1210212111102012100201102 |
4 | 13130011321222320011 |
5 | 31333414212042013 |
6 | 1030501520305445 |
7 | 51635005154444 |
oct | 7340571527005 |
9 | 1725442170642 |
10 | 511200112133 |
11 | 187887155517 |
12 | 830a7a72285 |
13 | 3928a66b769 |
14 | 1aa56904d5b |
15 | d46e04ed58 |
hex | 7705e6ae05 |
511200112133 has 2 divisors, whose sum is σ = 511200112134. Its totient is φ = 511200112132.
The previous prime is 511200112069. The next prime is 511200112183. The reversal of 511200112133 is 331211002115.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 494183692324 + 17016419809 = 702982^2 + 130447^2 .
It is a cyclic number.
It is not a de Polignac number, because 511200112133 - 26 = 511200112069 is a prime.
It is a super-2 number, since 2×5112001121332 (a number of 24 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (511200112183) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 255600056066 + 255600056067.
It is an arithmetic number, because the mean of its divisors is an integer number (255600056067).
Almost surely, 2511200112133 is an apocalyptic number.
It is an amenable number.
511200112133 is a deficient number, since it is larger than the sum of its proper divisors (1).
511200112133 is an equidigital number, since it uses as much as digits as its factorization.
511200112133 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 180, while the sum is 20.
Adding to 511200112133 its reverse (331211002115), we get a palindrome (842411114248).
The spelling of 511200112133 in words is "five hundred eleven billion, two hundred million, one hundred twelve thousand, one hundred thirty-three".
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