Base | Representation |
---|---|
bin | 10010110000000110… |
… | …011011010101001011 |
3 | 1220222012022122120212 |
4 | 102300012123111023 |
5 | 312213343002021 |
6 | 13125524511335 |
7 | 1311643046255 |
oct | 226006332513 |
9 | 56865278525 |
10 | 20134344011 |
11 | 85a2348731 |
12 | 3a9ab4b54b |
13 | 1b8b484732 |
14 | d90096cd5 |
15 | 7cc95685b |
hex | 4b019b54b |
20134344011 has 2 divisors, whose sum is σ = 20134344012. Its totient is φ = 20134344010.
The previous prime is 20134343981. The next prime is 20134344017. The reversal of 20134344011 is 11044343102.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 20134344011 - 222 = 20130149707 is a prime.
It is not a weakly prime, because it can be changed into another prime (20134344017) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10067172005 + 10067172006.
It is an arithmetic number, because the mean of its divisors is an integer number (10067172006).
Almost surely, 220134344011 is an apocalyptic number.
20134344011 is a deficient number, since it is larger than the sum of its proper divisors (1).
20134344011 is an equidigital number, since it uses as much as digits as its factorization.
20134344011 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1152, while the sum is 23.
Adding to 20134344011 its reverse (11044343102), we get a palindrome (31178687113).
Subtracting from 20134344011 its reverse (11044343102), we obtain a palindrome (9090000909).
The spelling of 20134344011 in words is "twenty billion, one hundred thirty-four million, three hundred forty-four thousand, eleven".
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