Base | Representation |
---|---|
bin | 10000001011101001110… |
… | …100001011111111111011 |
3 | 10221022022212222020111201 |
4 | 100023221310023333323 |
5 | 121204411212144311 |
6 | 2210505032435031 |
7 | 143224666611601 |
oct | 20135164137773 |
9 | 3838285866451 |
10 | 1112024334331 |
11 | 399674823389 |
12 | 15b626848a77 |
13 | 80b2bcaccc1 |
14 | 3bb7241b871 |
15 | 1ddd642b9c1 |
hex | 102e9d0bffb |
1112024334331 has 2 divisors, whose sum is σ = 1112024334332. Its totient is φ = 1112024334330.
The previous prime is 1112024334319. The next prime is 1112024334353. The reversal of 1112024334331 is 1334334202111.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1112024334331 - 235 = 1077664595963 is a prime.
It is a super-2 number, since 2×11120243343312 (a number of 25 digits) contains 22 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 1112024334331.
It is not a weakly prime, because it can be changed into another prime (1112024334131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 556012167165 + 556012167166.
It is an arithmetic number, because the mean of its divisors is an integer number (556012167166).
Almost surely, 21112024334331 is an apocalyptic number.
1112024334331 is a deficient number, since it is larger than the sum of its proper divisors (1).
1112024334331 is an equidigital number, since it uses as much as digits as its factorization.
1112024334331 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5184, while the sum is 28.
Adding to 1112024334331 its reverse (1334334202111), we get a palindrome (2446358536442).
The spelling of 1112024334331 in words is "one trillion, one hundred twelve billion, twenty-four million, three hundred thirty-four thousand, three hundred thirty-one".
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