Base | Representation |
---|---|
bin | 1010010101100100001011… |
… | …0110011100010101110011 |
3 | 1111020112121120000201001211 |
4 | 2211121002312130111303 |
5 | 2442203224342031311 |
6 | 40101141544140551 |
7 | 2252065200532513 |
oct | 245310266342563 |
9 | 44215546021054 |
10 | 11365605033331 |
11 | 369214281681a |
12 | 1336892167757 |
13 | 645a05caa2c3 |
14 | 2b4153725843 |
15 | 14a9a36d6821 |
hex | a5642d9c573 |
11365605033331 has 2 divisors, whose sum is σ = 11365605033332. Its totient is φ = 11365605033330.
The previous prime is 11365605033319. The next prime is 11365605033359. The reversal of 11365605033331 is 13333050656311.
11365605033331 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 11365605033331 - 213 = 11365605025139 is a prime.
It is a super-2 number, since 2×113656050333312 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (11365605033631) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5682802516665 + 5682802516666.
It is an arithmetic number, because the mean of its divisors is an integer number (5682802516666).
Almost surely, 211365605033331 is an apocalyptic number.
11365605033331 is a deficient number, since it is larger than the sum of its proper divisors (1).
11365605033331 is an equidigital number, since it uses as much as digits as its factorization.
11365605033331 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 218700, while the sum is 40.
Adding to 11365605033331 its reverse (13333050656311), we get a palindrome (24698655689642).
The spelling of 11365605033331 in words is "eleven trillion, three hundred sixty-five billion, six hundred five million, thirty-three thousand, three hundred thirty-one".
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