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2013101104211 is a prime number
BaseRepresentation
bin11101010010110110001…
…011010001110001010011
321010110011110212220022222
4131102312023101301103
5230440312340313321
64140450023051255
7265304344024121
oct35226613216123
97113143786288
102013101104211
1170682a10133a
1228619b5a0b2b
13117ab0875342
146d612816911
15375732513ab
hex1d4b62d1c53

2013101104211 has 2 divisors, whose sum is σ = 2013101104212. Its totient is φ = 2013101104210.

The previous prime is 2013101104181. The next prime is 2013101104213. The reversal of 2013101104211 is 1124011013102.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-2013101104211 is a prime.

Together with 2013101104213, it forms a pair of twin primes.

It is a Chen prime.

It is a self number, because there is not a number n which added to its sum of digits gives 2013101104211.

It is not a weakly prime, because it can be changed into another prime (2013101104213) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006550552105 + 1006550552106.

It is an arithmetic number, because the mean of its divisors is an integer number (1006550552106).

Almost surely, 22013101104211 is an apocalyptic number.

2013101104211 is a deficient number, since it is larger than the sum of its proper divisors (1).

2013101104211 is an equidigital number, since it uses as much as digits as its factorization.

2013101104211 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 48, while the sum is 17.

Adding to 2013101104211 its reverse (1124011013102), we get a palindrome (3137112117313).

The spelling of 2013101104211 in words is "two trillion, thirteen billion, one hundred one million, one hundred four thousand, two hundred eleven".