Base | Representation |
---|---|
bin | 1110001010000010000011… |
… | …00101000111110111011101 |
3 | 11002020002121112101222120121 |
4 | 13011001001211013313131 |
5 | 13040022313420240442 |
6 | 150113221123514541 |
7 | 6362066530151032 |
oct | 705010145076735 |
9 | 132202545358517 |
10 | 31131023211997 |
11 | 9a12658023425 |
12 | 35a9498828a51 |
13 | 144a848585ba7 |
14 | 798a6cb2d389 |
15 | 38ebc911c967 |
hex | 1c5041947ddd |
31131023211997 has 2 divisors, whose sum is σ = 31131023211998. Its totient is φ = 31131023211996.
The previous prime is 31131023211989. The next prime is 31131023212057. The reversal of 31131023211997 is 79911232013113.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 29471283132516 + 1659740079481 = 5428746^2 + 1288309^2 .
It is a cyclic number.
It is not a de Polignac number, because 31131023211997 - 23 = 31131023211989 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (31131023211697) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15565511605998 + 15565511605999.
It is an arithmetic number, because the mean of its divisors is an integer number (15565511605999).
Almost surely, 231131023211997 is an apocalyptic number.
It is an amenable number.
31131023211997 is a deficient number, since it is larger than the sum of its proper divisors (1).
31131023211997 is an equidigital number, since it uses as much as digits as its factorization.
31131023211997 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 61236, while the sum is 43.
The spelling of 31131023211997 in words is "thirty-one trillion, one hundred thirty-one billion, twenty-three million, two hundred eleven thousand, nine hundred ninety-seven".
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