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2013112199981 is a prime number
BaseRepresentation
bin11101010010110110110…
…101100110101100101101
321010110012020202121212102
4131102312311212230231
5230440323210344411
64140451100544445
7265304540236241
oct35226665465455
97113166677772
102013112199981
1170683539a7a3
122861a3252125
13117ab2c5b89c
146d6140a4421
15375741e3d3b
hex1d4b6d66b2d

2013112199981 has 2 divisors, whose sum is σ = 2013112199982. Its totient is φ = 2013112199980.

The previous prime is 2013112199957. The next prime is 2013112200097. The reversal of 2013112199981 is 1899912113102.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1554422965225 + 458689234756 = 1246765^2 + 677266^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-2013112199981 is a prime.

It is a super-2 number, since 2×20131121999812 (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (2013112199911) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006556099990 + 1006556099991.

It is an arithmetic number, because the mean of its divisors is an integer number (1006556099991).

Almost surely, 22013112199981 is an apocalyptic number.

It is an amenable number.

2013112199981 is a deficient number, since it is larger than the sum of its proper divisors (1).

2013112199981 is an equidigital number, since it uses as much as digits as its factorization.

2013112199981 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 69984, while the sum is 47.

The spelling of 2013112199981 in words is "two trillion, thirteen billion, one hundred twelve million, one hundred ninety-nine thousand, nine hundred eighty-one".