Base | Representation |
---|---|
bin | 11101110101001110100101… |
… | …011110000001000000001111 |
3 | 122012112201002011210002101122 |
4 | 131311032211132001000033 |
5 | 114144044414334144413 |
6 | 1143012550025431155 |
7 | 36430653416123204 |
oct | 3565164536010017 |
9 | 565481064702348 |
10 | 131201142099983 |
11 | 388941013439aa |
12 | 1286b7b103b4bb |
13 | 58292a36711a7 |
14 | 2458252263cab |
15 | 1027c99684108 |
hex | 7753a578100f |
131201142099983 has 2 divisors, whose sum is σ = 131201142099984. Its totient is φ = 131201142099982.
The previous prime is 131201142099931. The next prime is 131201142100027. The reversal of 131201142099983 is 389990241102131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 131201142099983 - 218 = 131201141837839 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131201142092983) 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, 65600571049991 + 65600571049992.
It is an arithmetic number, because the mean of its divisors is an integer number (65600571049992).
Almost surely, 2131201142099983 is an apocalyptic number.
131201142099983 is a deficient number, since it is larger than the sum of its proper divisors (1).
131201142099983 is an equidigital number, since it uses as much as digits as its factorization.
131201142099983 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 839808, while the sum is 53.
The spelling of 131201142099983 in words is "one hundred thirty-one trillion, two hundred one billion, one hundred forty-two million, ninety-nine thousand, nine hundred eighty-three".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.075 sec. • engine limits •