Base | Representation |
---|---|
bin | 1011111100011101111010… |
… | …0110000011011000001011 |
3 | 1201111112201200100121201221 |
4 | 2333013132212003120023 |
5 | 3210134301040304011 |
6 | 43533231243222511 |
7 | 2523601060110103 |
oct | 277073646033013 |
9 | 51445650317657 |
10 | 13133449541131 |
11 | 4203956972167 |
12 | 1581425817a37 |
13 | 74362c41033a |
14 | 33593b6b6203 |
15 | 17b97028b971 |
hex | bf1de98360b |
13133449541131 has 2 divisors, whose sum is σ = 13133449541132. Its totient is φ = 13133449541130.
The previous prime is 13133449541101. The next prime is 13133449541161. The reversal of 13133449541131 is 13114594433131.
It is a balanced prime because it is at equal distance from previous prime (13133449541101) and next prime (13133449541161).
It is a cyclic number.
It is not a de Polignac number, because 13133449541131 - 25 = 13133449541099 is a prime.
It is a super-2 number, since 2×131334495411312 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (13133449541101) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6566724770565 + 6566724770566.
It is an arithmetic number, because the mean of its divisors is an integer number (6566724770566).
Almost surely, 213133449541131 is an apocalyptic number.
13133449541131 is a deficient number, since it is larger than the sum of its proper divisors (1).
13133449541131 is an equidigital number, since it uses as much as digits as its factorization.
13133449541131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 233280, while the sum is 43.
The spelling of 13133449541131 in words is "thirteen trillion, one hundred thirty-three billion, four hundred forty-nine million, five hundred forty-one thousand, one hundred thirty-one".
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