Base | Representation |
---|---|
bin | 11101110101001110100101… |
… | …011100101011000010010001 |
3 | 122012112201002011010012100221 |
4 | 131311032211130223002101 |
5 | 114144044414241412412 |
6 | 1143012550014113041 |
7 | 36430653413125456 |
oct | 3565164534530221 |
9 | 565481064105327 |
10 | 131201141747857 |
11 | 38894101123395 |
12 | 1286b7b0aab781 |
13 | 58292a3579b2c |
14 | 24582521b182d |
15 | 1027c99614b07 |
hex | 7753a572b091 |
131201141747857 has 2 divisors, whose sum is σ = 131201141747858. Its totient is φ = 131201141747856.
The previous prime is 131201141747843. The next prime is 131201141747867. The reversal of 131201141747857 is 758747141102131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 130523221704976 + 677920042881 = 11424676^2 + 823359^2 .
It is a cyclic number.
It is not a de Polignac number, because 131201141747857 - 231 = 131198994264209 is a prime.
It is not a weakly prime, because it can be changed into another prime (131201141747867) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65600570873928 + 65600570873929.
It is an arithmetic number, because the mean of its divisors is an integer number (65600570873929).
Almost surely, 2131201141747857 is an apocalyptic number.
It is an amenable number.
131201141747857 is a deficient number, since it is larger than the sum of its proper divisors (1).
131201141747857 is an equidigital number, since it uses as much as digits as its factorization.
131201141747857 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1317120, while the sum is 52.
Adding to 131201141747857 its reverse (758747141102131), we get a palindrome (889948282849988).
The spelling of 131201141747857 in words is "one hundred thirty-one trillion, two hundred one billion, one hundred forty-one million, seven hundred forty-seven thousand, eight hundred fifty-seven".
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