Base | Representation |
---|---|
bin | 1011111011111111111100… |
… | …0100011010000000000011 |
3 | 1201110202222000001212002021 |
4 | 2332333333010122000003 |
5 | 3210021312013400220 |
6 | 43525421052502311 |
7 | 2523164520612406 |
oct | 276777704320003 |
9 | 51422860055067 |
10 | 13125404434435 |
11 | 42004a8696457 |
12 | 157b95b47a997 |
13 | 742949736187 |
14 | 3353b7029d3d |
15 | 17b64dd314aa |
hex | befff11a003 |
13125404434435 has 16 divisors (see below), whose sum is σ = 15771897319680. Its totient is φ = 10486053658080.
The previous prime is 13125404434427. The next prime is 13125404434459. The reversal of 13125404434435 is 53443440452131.
It is not a de Polignac number, because 13125404434435 - 23 = 13125404434427 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13125404434391 and 13125404434400.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 21913954 + ... + 22504936.
It is an arithmetic number, because the mean of its divisors is an integer number (985743582480).
Almost surely, 213125404434435 is an apocalyptic number.
13125404434435 is a deficient number, since it is larger than the sum of its proper divisors (2646492885245).
13125404434435 is an equidigital number, since it uses as much as digits as its factorization.
13125404434435 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 597018.
The product of its (nonzero) digits is 1382400, while the sum is 43.
Adding to 13125404434435 its reverse (53443440452131), we get a palindrome (66568844886566).
The spelling of 13125404434435 in words is "thirteen trillion, one hundred twenty-five billion, four hundred four million, four hundred thirty-four thousand, four hundred thirty-five".
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