Base | Representation |
---|---|
bin | 11000001001110011010010… |
… | …01100110010001001001011 |
3 | 20222001120210022211112111021 |
4 | 30010321221030302021023 |
5 | 23430202100104304011 |
6 | 304543532135413311 |
7 | 14121211562402140 |
oct | 1404715114621113 |
9 | 228046708745437 |
10 | 53113330541131 |
11 | 15a18267479803 |
12 | 5b59876b24237 |
13 | 2383747766473 |
14 | d189adb703c7 |
15 | 6218eda83471 |
hex | 304e6933224b |
53113330541131 has 8 divisors (see below), whose sum is σ = 64271593260000. Its totient is φ = 42847728839808.
The previous prime is 53113330541099. The next prime is 53113330541143. The reversal of 53113330541131 is 13114503331135.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 53113330541131 - 25 = 53113330541099 is a prime.
It is a super-2 number, since 2×531133305411312 (a number of 28 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (53113330542131) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 223165254256 + ... + 223165254493.
It is an arithmetic number, because the mean of its divisors is an integer number (8033949157500).
Almost surely, 253113330541131 is an apocalyptic number.
53113330541131 is a deficient number, since it is larger than the sum of its proper divisors (11158262718869).
53113330541131 is a wasteful number, since it uses less digits than its factorization.
53113330541131 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 446330508773.
The product of its (nonzero) digits is 24300, while the sum is 34.
Adding to 53113330541131 its reverse (13114503331135), we get a palindrome (66227833872266).
The spelling of 53113330541131 in words is "fifty-three trillion, one hundred thirteen billion, three hundred thirty million, five hundred forty-one thousand, one hundred thirty-one".
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