Base | Representation |
---|---|
bin | 1110001100000110010011… |
… | …11011000001100010000011 |
3 | 11002110212220010110020002212 |
4 | 13012003021323001202003 |
5 | 13042203224203200311 |
6 | 150210003104332335 |
7 | 6400162252050413 |
oct | 706031173014203 |
9 | 132425803406085 |
10 | 31202030131331 |
11 | 9a3a785634934 |
12 | 35bb1b49140ab |
13 | 14544534c0ca7 |
14 | 79c287374643 |
15 | 391982d29c8b |
hex | 1c60c9ec1883 |
31202030131331 has 4 divisors (see below), whose sum is σ = 31240598772600. Its totient is φ = 31163461490064.
The previous prime is 31202030131301. The next prime is 31202030131333. The reversal of 31202030131331 is 13313103020213.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-31202030131331 is a prime.
It is a super-4 number, since 4×312020301313314 (a number of 55 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 31202030131297 and 31202030131306.
It is not an unprimeable number, because it can be changed into a prime (31202030131333) 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, 19284319421 + ... + 19284321038.
It is an arithmetic number, because the mean of its divisors is an integer number (7810149693150).
Almost surely, 231202030131331 is an apocalyptic number.
31202030131331 is a deficient number, since it is larger than the sum of its proper divisors (38568641269).
31202030131331 is an equidigital number, since it uses as much as digits as its factorization.
31202030131331 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 38568641268.
The product of its (nonzero) digits is 972, while the sum is 23.
Adding to 31202030131331 its reverse (13313103020213), we get a palindrome (44515133151544).
The spelling of 31202030131331 in words is "thirty-one trillion, two hundred two billion, thirty million, one hundred thirty-one thousand, three hundred thirty-one".
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