Base | Representation |
---|---|
bin | 1110001100101010001110… |
… | …01101100011111100001011 |
3 | 11002112201122220201020212221 |
4 | 13012111013031203330023 |
5 | 13043012221333341133 |
6 | 150222512302114511 |
7 | 6401444134246156 |
oct | 706250715437413 |
9 | 132481586636787 |
10 | 31221312012043 |
11 | 9a4797a76923a |
12 | 3602a9627ba37 |
13 | 145620718416a |
14 | 79d196126c9d |
15 | 3922109e4a2d |
hex | 1c6547363f0b |
31221312012043 has 2 divisors, whose sum is σ = 31221312012044. Its totient is φ = 31221312012042.
The previous prime is 31221312012031. The next prime is 31221312012083. The reversal of 31221312012043 is 34021021312213.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-31221312012043 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 31221312011999 and 31221312012017.
It is not a weakly prime, because it can be changed into another prime (31221312012013) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15610656006021 + 15610656006022.
It is an arithmetic number, because the mean of its divisors is an integer number (15610656006022).
Almost surely, 231221312012043 is an apocalyptic number.
31221312012043 is a deficient number, since it is larger than the sum of its proper divisors (1).
31221312012043 is an equidigital number, since it uses as much as digits as its factorization.
31221312012043 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1728, while the sum is 25.
Adding to 31221312012043 its reverse (34021021312213), we get a palindrome (65242333324256).
The spelling of 31221312012043 in words is "thirty-one trillion, two hundred twenty-one billion, three hundred twelve million, twelve thousand, forty-three".
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