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13011313033307 is a prime number
BaseRepresentation
bin1011110101010110111010…
…1100010000000001011011
31201001212110202222011111002
42331111232230100001123
53201134132114031212
643401151550055215
72512015315655651
oct275255654200133
951055422864432
1013011313033307
1141670810040a6
12156181a4aa50b
13734c666c4c74
1432da726dcdd1
151786c295bac2
hexbd56eb1005b

13011313033307 has 2 divisors, whose sum is σ = 13011313033308. Its totient is φ = 13011313033306.

The previous prime is 13011313033289. The next prime is 13011313033421. The reversal of 13011313033307 is 70333031311031.

13011313033307 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 13011313033307 - 228 = 13011044597851 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 13011313033307.

It is not a weakly prime, because it can be changed into another prime (13011313035307) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6505656516653 + 6505656516654.

It is an arithmetic number, because the mean of its divisors is an integer number (6505656516654).

Almost surely, 213011313033307 is an apocalyptic number.

13011313033307 is a deficient number, since it is larger than the sum of its proper divisors (1).

13011313033307 is an equidigital number, since it uses as much as digits as its factorization.

13011313033307 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 5103, while the sum is 29.

Adding to 13011313033307 its reverse (70333031311031), we get a palindrome (83344344344338).

The spelling of 13011313033307 in words is "thirteen trillion, eleven billion, three hundred thirteen million, thirty-three thousand, three hundred seven".