Base | Representation |
---|---|
bin | 11101100101100001110010… |
… | …000101111000100000011011 |
3 | 122001201120020212102121021022 |
4 | 131211201302011320200123 |
5 | 114023410323244010121 |
6 | 1140425200001033055 |
7 | 36260010331040543 |
oct | 3545416205704033 |
9 | 561646225377238 |
10 | 130122243344411 |
11 | 385085960a719a |
12 | 1271669002578b |
13 | 577b623468a86 |
14 | 241bd43426323 |
15 | 1009ba147daab |
hex | 76587217881b |
130122243344411 has 2 divisors, whose sum is σ = 130122243344412. Its totient is φ = 130122243344410.
The previous prime is 130122243344389. The next prime is 130122243344473. The reversal of 130122243344411 is 114443342221031.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 130122243344411 - 210 = 130122243343387 is a prime.
It is a super-2 number, since 2×1301222433444112 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (130122243344491) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65061121672205 + 65061121672206.
It is an arithmetic number, because the mean of its divisors is an integer number (65061121672206).
Almost surely, 2130122243344411 is an apocalyptic number.
130122243344411 is a deficient number, since it is larger than the sum of its proper divisors (1).
130122243344411 is an equidigital number, since it uses as much as digits as its factorization.
130122243344411 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 55296, while the sum is 35.
Adding to 130122243344411 its reverse (114443342221031), we get a palindrome (244565585565442).
The spelling of 130122243344411 in words is "one hundred thirty trillion, one hundred twenty-two billion, two hundred forty-three million, three hundred forty-four thousand, four hundred eleven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.071 sec. • engine limits •