Base | Representation |
---|---|
bin | 111101010001000111… |
… | …0101011010111000011 |
3 | 110120121101121220111212 |
4 | 1322202032223113003 |
5 | 4123424104303431 |
6 | 140235350200335 |
7 | 12335306065511 |
oct | 1724216532703 |
9 | 416541556455 |
10 | 131570775491 |
11 | 508872aaaa1 |
12 | 215ba9530ab |
13 | c53a44759c |
14 | 6521d534b1 |
15 | 3650bdd72b |
hex | 1ea23ab5c3 |
131570775491 has 2 divisors, whose sum is σ = 131570775492. Its totient is φ = 131570775490.
The previous prime is 131570775481. The next prime is 131570775493. The reversal of 131570775491 is 194577075131.
It is a strong prime.
It is an emirp because it is prime and its reverse (194577075131) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-131570775491 is a prime.
It is a Sophie Germain prime.
Together with 131570775493, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (131570775493) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65785387745 + 65785387746.
It is an arithmetic number, because the mean of its divisors is an integer number (65785387746).
Almost surely, 2131570775491 is an apocalyptic number.
131570775491 is a deficient number, since it is larger than the sum of its proper divisors (1).
131570775491 is an equidigital number, since it uses as much as digits as its factorization.
131570775491 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 926100, while the sum is 50.
The spelling of 131570775491 in words is "one hundred thirty-one billion, five hundred seventy million, seven hundred seventy-five thousand, four hundred ninety-one".
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