Base | Representation |
---|---|
bin | 1010111011010000010010… |
… | …1111100011100100100011 |
3 | 1120112102220210212122221112 |
4 | 2232310010233203210203 |
5 | 3033310313343232433 |
6 | 41314424323013535 |
7 | 2346626615362535 |
oct | 256640457434443 |
9 | 46472823778845 |
10 | 12013103102243 |
11 | 391180157a521 |
12 | 1420277b0a2ab |
13 | 691aa5250119 |
14 | 2d7619d05855 |
15 | 15c74d43b348 |
hex | aed04be3923 |
12013103102243 has 2 divisors, whose sum is σ = 12013103102244. Its totient is φ = 12013103102242.
The previous prime is 12013103102131. The next prime is 12013103102281. The reversal of 12013103102243 is 34220130131021.
12013103102243 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 12013103102243 - 224 = 12013086325027 is a prime.
It is a super-2 number, since 2×120131031022432 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (12013103101243) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6006551551121 + 6006551551122.
It is an arithmetic number, because the mean of its divisors is an integer number (6006551551122).
Almost surely, 212013103102243 is an apocalyptic number.
12013103102243 is a deficient number, since it is larger than the sum of its proper divisors (1).
12013103102243 is an equidigital number, since it uses as much as digits as its factorization.
12013103102243 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 864, while the sum is 23.
Adding to 12013103102243 its reverse (34220130131021), we get a palindrome (46233233233264).
The spelling of 12013103102243 in words is "twelve trillion, thirteen billion, one hundred three million, one hundred two thousand, two hundred forty-three".
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