Base | Representation |
---|---|
bin | 1100000000011010001000… |
… | …0111000010101000101011 |
3 | 1201202000111010201110112000 |
4 | 3000012202013002220223 |
5 | 3212241430442133103 |
6 | 44024305422222043 |
7 | 2531515623511452 |
oct | 300064207025053 |
9 | 51660433643460 |
10 | 13201154255403 |
11 | 422a63a473084 |
12 | 159257ba24923 |
13 | 749b2c17354a |
14 | 338d21529b99 |
15 | 17d5d410caa3 |
hex | c01a21c2a2b |
13201154255403 has 16 divisors (see below), whose sum is σ = 20034272042400. Its totient is φ = 8586116588160.
The previous prime is 13201154255389. The next prime is 13201154255411. The reversal of 13201154255403 is 30455245110231.
13201154255403 is a `hidden beast` number, since 1 + 3 + 20 + 1 + 1 + 542 + 55 + 40 + 3 = 666.
It is not a de Polignac number, because 13201154255403 - 25 = 13201154255371 is a prime.
It is not an unprimeable number, because it can be changed into a prime (13201154255483) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 5962579858 + ... + 5962582071.
It is an arithmetic number, because the mean of its divisors is an integer number (1252142002650).
Almost surely, 213201154255403 is an apocalyptic number.
13201154255403 is a deficient number, since it is larger than the sum of its proper divisors (6833117786997).
13201154255403 is a wasteful number, since it uses less digits than its factorization.
13201154255403 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 11925161979 (or 11925161973 counting only the distinct ones).
The product of its (nonzero) digits is 72000, while the sum is 36.
Adding to 13201154255403 its reverse (30455245110231), we get a palindrome (43656399365634).
The spelling of 13201154255403 in words is "thirteen trillion, two hundred one billion, one hundred fifty-four million, two hundred fifty-five thousand, four hundred three".
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