Base | Representation |
---|---|
bin | 1100000110110011110111… |
… | …0001011001101111111001 |
3 | 1202010112101020022202020122 |
4 | 3001230331301121233321 |
5 | 3221042203320323001 |
6 | 44151015302503025 |
7 | 2542461313144160 |
oct | 301547561315771 |
9 | 52115336282218 |
10 | 13311140011001 |
11 | 427224a574158 |
12 | 15ab955a34a75 |
13 | 7573097239c1 |
14 | 3403968823d7 |
15 | 1813bed3eb1b |
hex | c1b3dc59bf9 |
13311140011001 has 4 divisors (see below), whose sum is σ = 15212731441152. Its totient is φ = 11409548580852.
The previous prime is 13311140010977. The next prime is 13311140011003. The reversal of 13311140011001 is 10011004111331.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13311140011001 is a prime.
It is not an unprimeable number, because it can be changed into a prime (13311140011003) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 950795715065 + ... + 950795715078.
It is an arithmetic number, because the mean of its divisors is an integer number (3803182860288).
Almost surely, 213311140011001 is an apocalyptic number.
It is an amenable number.
13311140011001 is a deficient number, since it is larger than the sum of its proper divisors (1901591430151).
13311140011001 is an equidigital number, since it uses as much as digits as its factorization.
13311140011001 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1901591430150.
The product of its (nonzero) digits is 36, while the sum is 17.
Adding to 13311140011001 its reverse (10011004111331), we get a palindrome (23322144122332).
The spelling of 13311140011001 in words is "thirteen trillion, three hundred eleven billion, one hundred forty million, eleven thousand, one".
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