Base | Representation |
---|---|
bin | 1111011101110100000… |
… | …11011100100100000011 |
3 | 1212210122100110002202101 |
4 | 13232322003130210003 |
5 | 32201303011134021 |
6 | 1044042314331231 |
7 | 53251435144336 |
oct | 7567203344403 |
9 | 1783570402671 |
10 | 531402443011 |
11 | 195403909740 |
12 | 86ba5729b17 |
13 | 3b159c385bb |
14 | 1ba11a1861d |
15 | dc52935a91 |
hex | 7bba0dc903 |
531402443011 has 4 divisors (see below), whose sum is σ = 579711756024. Its totient is φ = 483093130000.
The previous prime is 531402443009. The next prime is 531402443107. The reversal of 531402443011 is 110344204135.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 531402443011 - 21 = 531402443009 is a prime.
It is a super-2 number, since 2×5314024430112 (a number of 24 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (531402444011) 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, 24154656490 + ... + 24154656511.
It is an arithmetic number, because the mean of its divisors is an integer number (144927939006).
Almost surely, 2531402443011 is an apocalyptic number.
531402443011 is a deficient number, since it is larger than the sum of its proper divisors (48309313013).
531402443011 is a wasteful number, since it uses less digits than its factorization.
531402443011 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 48309313012.
The product of its (nonzero) digits is 5760, while the sum is 28.
Adding to 531402443011 its reverse (110344204135), we get a palindrome (641746647146).
The spelling of 531402443011 in words is "five hundred thirty-one billion, four hundred two million, four hundred forty-three thousand, eleven".
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