Base | Representation |
---|---|
bin | 1001001001110111010000… |
… | …00111111011010011100001 |
3 | 2122021102100001112022012012 |
4 | 10210323220013323103201 |
5 | 10114302431423003324 |
6 | 110451351223242305 |
7 | 4145231630332505 |
oct | 444735007732341 |
9 | 78242301468165 |
10 | 20130111141089 |
11 | 646114aa48104 |
12 | 23114255a9395 |
13 | b30352200111 |
14 | 4d84343b8305 |
15 | 24d96c85020e |
hex | 124ee81fb4e1 |
20130111141089 has 2 divisors, whose sum is σ = 20130111141090. Its totient is φ = 20130111141088.
The previous prime is 20130111141083. The next prime is 20130111141103. The reversal of 20130111141089 is 98014111103102.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 19649006598400 + 481104542689 = 4432720^2 + 693617^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-20130111141089 is a prime.
It is not a weakly prime, because it can be changed into another prime (20130111141083) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10065055570544 + 10065055570545.
It is an arithmetic number, because the mean of its divisors is an integer number (10065055570545).
Almost surely, 220130111141089 is an apocalyptic number.
It is an amenable number.
20130111141089 is a deficient number, since it is larger than the sum of its proper divisors (1).
20130111141089 is an equidigital number, since it uses as much as digits as its factorization.
20130111141089 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1728, while the sum is 32.
The spelling of 20130111141089 in words is "twenty trillion, one hundred thirty billion, one hundred eleven million, one hundred forty-one thousand, eighty-nine".
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