Base | Representation |
---|---|
bin | 1010010011100100000101… |
… | …1101100101100111010011 |
3 | 1111010020212100021202012211 |
4 | 2210321001131211213103 |
5 | 2441122320313032011 |
6 | 40033254100332551 |
7 | 2246440146561163 |
oct | 244710135454723 |
9 | 44106770252184 |
10 | 11331222002131 |
11 | 36795a9463559 |
12 | 13300973a4157 |
13 | 6426b6843a25 |
14 | 2b261113c5a3 |
15 | 149b3edc6b21 |
hex | a4e417659d3 |
11331222002131 has 2 divisors, whose sum is σ = 11331222002132. Its totient is φ = 11331222002130.
The previous prime is 11331222002101. The next prime is 11331222002143. The reversal of 11331222002131 is 13120022213311.
11331222002131 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 11331222002131 - 217 = 11331221871059 is a prime.
It is a super-2 number, since 2×113312220021312 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (11331222002101) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5665611001065 + 5665611001066.
It is an arithmetic number, because the mean of its divisors is an integer number (5665611001066).
Almost surely, 211331222002131 is an apocalyptic number.
11331222002131 is a deficient number, since it is larger than the sum of its proper divisors (1).
11331222002131 is an equidigital number, since it uses as much as digits as its factorization.
11331222002131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 432, while the sum is 22.
Adding to 11331222002131 its reverse (13120022213311), we get a palindrome (24451244215442).
The spelling of 11331222002131 in words is "eleven trillion, three hundred thirty-one billion, two hundred twenty-two million, two thousand, one hundred thirty-one".
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