Base | Representation |
---|---|
bin | 1011110100110001000100… |
… | …0110110011001001100011 |
3 | 1201000220020201001222222111 |
4 | 2331030101012303021203 |
5 | 3201002330132000120 |
6 | 43352351445220151 |
7 | 2511205456055464 |
oct | 275142106631143 |
9 | 51026221058874 |
10 | 13001153000035 |
11 | 4162837a24792 |
12 | 155b863a01657 |
13 | 734007804b65 |
14 | 32d38b1dcd6b |
15 | 1782ca9db75a |
hex | bd3111b3263 |
13001153000035 has 16 divisors (see below), whose sum is σ = 15850279821600. Its totient is φ = 10236189643776.
The previous prime is 13001153000027. The next prime is 13001153000057. The reversal of 13001153000035 is 53000035110031.
It is a cyclic number.
It is not a de Polignac number, because 13001153000035 - 23 = 13001153000027 is a prime.
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, 74791680 + ... + 74965309.
It is an arithmetic number, because the mean of its divisors is an integer number (990642488850).
Almost surely, 213001153000035 is an apocalyptic number.
13001153000035 is a deficient number, since it is larger than the sum of its proper divisors (2849126821565).
13001153000035 is a wasteful number, since it uses less digits than its factorization.
13001153000035 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 149757270.
The product of its (nonzero) digits is 675, while the sum is 22.
Adding to 13001153000035 its reverse (53000035110031), we get a palindrome (66001188110066).
The spelling of 13001153000035 in words is "thirteen trillion, one billion, one hundred fifty-three million, thirty-five", and thus it is an aban number.
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