Base | Representation |
---|---|
bin | 1010010010011001111010… |
… | …1100010010110111110101 |
3 | 1111001100110201210101112120 |
4 | 2210212132230102313311 |
5 | 2440311031143132013 |
6 | 40020202241110153 |
7 | 2245133560453143 |
oct | 244463654226765 |
9 | 44040421711476 |
10 | 11311311302133 |
11 | 3671111387997 |
12 | 132825b328359 |
13 | 64086280453b |
14 | 2b1682a60593 |
15 | 149376dea123 |
hex | a499eb12df5 |
11311311302133 has 32 divisors (see below), whose sum is σ = 15821424423936. Its totient is φ = 7174545404928.
The previous prime is 11311311302111. The next prime is 11311311302183. The reversal of 11311311302133 is 33120311311311.
It is not a de Polignac number, because 11311311302133 - 26 = 11311311302069 is a prime.
It is a super-3 number, since 3×113113113021333 (a number of 40 digits) contains 333 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (11311311302183) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 92452066 + ... + 92574332.
It is an arithmetic number, because the mean of its divisors is an integer number (494419513248).
Almost surely, 211311311302133 is an apocalyptic number.
It is an amenable number.
11311311302133 is a deficient number, since it is larger than the sum of its proper divisors (4510113121803).
11311311302133 is a wasteful number, since it uses less digits than its factorization.
11311311302133 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 129433.
The product of its (nonzero) digits is 486, while the sum is 24.
Adding to 11311311302133 its reverse (33120311311311), we get a palindrome (44431622613444).
The spelling of 11311311302133 in words is "eleven trillion, three hundred eleven billion, three hundred eleven million, three hundred two thousand, one hundred thirty-three".
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