Base | Representation |
---|---|
bin | 11010001101001011100111… |
… | …101111001011011011001011 |
3 | 120010002021101221000211011012 |
4 | 122031023213233023123023 |
5 | 110101314310310402243 |
6 | 1045043233511354135 |
7 | 33163621225340543 |
oct | 3215134757133313 |
9 | 503067357024135 |
10 | 115255040325323 |
11 | 337a642a630a18 |
12 | 10b1524798294b |
13 | 4c40671731b12 |
14 | 2066533696323 |
15 | d4d0ac4bcd18 |
hex | 68d2e7bcb6cb |
115255040325323 has 2 divisors, whose sum is σ = 115255040325324. Its totient is φ = 115255040325322.
The previous prime is 115255040325287. The next prime is 115255040325329. The reversal of 115255040325323 is 323523040552511.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 115255040325323 - 218 = 115255040063179 is a prime.
It is a super-3 number, since 3×1152550403253233 (a number of 43 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (115255040325329) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57627520162661 + 57627520162662.
It is an arithmetic number, because the mean of its divisors is an integer number (57627520162662).
Almost surely, 2115255040325323 is an apocalyptic number.
115255040325323 is a deficient number, since it is larger than the sum of its proper divisors (1).
115255040325323 is an equidigital number, since it uses as much as digits as its factorization.
115255040325323 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 540000, while the sum is 41.
Adding to 115255040325323 its reverse (323523040552511), we get a palindrome (438778080877834).
The spelling of 115255040325323 in words is "one hundred fifteen trillion, two hundred fifty-five billion, forty million, three hundred twenty-five thousand, three hundred twenty-three".
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