Search a number
-
+
20130011313 = 36710003771
BaseRepresentation
bin10010101111110101…
…111001100010110001
31220221220011112020120
4102233311321202301
5312211240330223
613125252000453
71311561162426
oct225765714261
956856145216
1020130011313
11859a959495
123a99600129
131b8a5c85c9
14d8d689d4d
157cc39cbe3
hex4afd798b1

20130011313 has 4 divisors (see below), whose sum is σ = 26840015088. Its totient is φ = 13420007540.

The previous prime is 20130011281. The next prime is 20130011383. The reversal of 20130011313 is 31311003102.

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 not a de Polignac number, because 20130011313 - 25 = 20130011281 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 20130011292 and 20130011301.

It is not an unprimeable number, because it can be changed into a prime (20130011383) 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, 3355001883 + ... + 3355001888.

It is an arithmetic number, because the mean of its divisors is an integer number (6710003772).

Almost surely, 220130011313 is an apocalyptic number.

It is an amenable number.

20130011313 is a deficient number, since it is larger than the sum of its proper divisors (6710003775).

20130011313 is an equidigital number, since it uses as much as digits as its factorization.

20130011313 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 6710003774.

The product of its (nonzero) digits is 54, while the sum is 15.

Adding to 20130011313 its reverse (31311003102), we get a palindrome (51441014415).

The spelling of 20130011313 in words is "twenty billion, one hundred thirty million, eleven thousand, three hundred thirteen".

Divisors: 1 3 6710003771 20130011313