Base | Representation |
---|---|
bin | 1110001001011011000110… |
… | …01111000010001010011011 |
3 | 11002011002122021011200101001 |
4 | 13010231203033002022123 |
5 | 13034202002412312311 |
6 | 150055442201405431 |
7 | 6360426413315665 |
oct | 704554317021233 |
9 | 132132567150331 |
10 | 31110113010331 |
11 | 9a047a7834423 |
12 | 35a5421a99877 |
13 | 14488954522b4 |
14 | 797a479d3535 |
15 | 38e3a350c4c1 |
hex | 1c4b633c229b |
31110113010331 has 2 divisors, whose sum is σ = 31110113010332. Its totient is φ = 31110113010330.
The previous prime is 31110113010299. The next prime is 31110113010341. The reversal of 31110113010331 is 13301031101113.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 31110113010331 - 25 = 31110113010299 is a prime.
It is a super-2 number, since 2×311101130103312 (a number of 28 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 31110113010299 and 31110113010308.
It is not a weakly prime, because it can be changed into another prime (31110113010341) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15555056505165 + 15555056505166.
It is an arithmetic number, because the mean of its divisors is an integer number (15555056505166).
Almost surely, 231110113010331 is an apocalyptic number.
31110113010331 is a deficient number, since it is larger than the sum of its proper divisors (1).
31110113010331 is an equidigital number, since it uses as much as digits as its factorization.
31110113010331 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 81, while the sum is 19.
Adding to 31110113010331 its reverse (13301031101113), we get a palindrome (44411144111444).
The spelling of 31110113010331 in words is "thirty-one trillion, one hundred ten billion, one hundred thirteen million, ten thousand, three hundred thirty-one".
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