Base | Representation |
---|---|
bin | 100101001100001001101… |
… | …0000110000001101101011 |
3 | 200002122020200010120112110 |
4 | 1022120103100300031223 |
5 | 1132221003031204011 |
6 | 14512040423413403 |
7 | 1035165425124336 |
oct | 112302320601553 |
9 | 20078220116473 |
10 | 5111334241131 |
11 | 16a0782a21798 |
12 | 6a673b5b4263 |
13 | 2b0cc5b27661 |
14 | 139565c4161d |
15 | 8ce56a1eba6 |
hex | 4a61343036b |
5111334241131 has 4 divisors (see below), whose sum is σ = 6815112321512. Its totient is φ = 3407556160752.
The previous prime is 5111334241127. The next prime is 5111334241147. The reversal of 5111334241131 is 1311424331115.
It is a happy number.
It is a semiprime because it is the product of two primes.
It is not a de Polignac number, because 5111334241131 - 22 = 5111334241127 is a prime.
It is a super-2 number, since 2×51113342411312 (a number of 26 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (5111334241031) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 851889040186 + ... + 851889040191.
It is an arithmetic number, because the mean of its divisors is an integer number (1703778080378).
Almost surely, 25111334241131 is an apocalyptic number.
5111334241131 is a deficient number, since it is larger than the sum of its proper divisors (1703778080381).
5111334241131 is a wasteful number, since it uses less digits than its factorization.
5111334241131 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1703778080380.
The product of its digits is 4320, while the sum is 30.
Adding to 5111334241131 its reverse (1311424331115), we get a palindrome (6422758572246).
The spelling of 5111334241131 in words is "five trillion, one hundred eleven billion, three hundred thirty-four million, two hundred forty-one thousand, one hundred thirty-one".
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