Base | Representation |
---|---|
bin | 1100000110001011100101… |
… | …1100111100001100111010 |
3 | 1202002111110212121101112001 |
4 | 3001202321130330030322 |
5 | 3220403034002314434 |
6 | 44142030505211214 |
7 | 2541626402452021 |
oct | 301427134741472 |
9 | 52074425541461 |
10 | 13300330119994 |
11 | 4268702683035 |
12 | 15a9839790b0a |
13 | 7562a5ccc5a8 |
14 | 33da4cd3b8b8 |
15 | 180e8ad04114 |
hex | c18b973c33a |
13300330119994 has 4 divisors (see below), whose sum is σ = 19950495179994. Its totient is φ = 6650165059996.
The previous prime is 13300330119959. The next prime is 13300330120043. The reversal of 13300330119994 is 49991103300331.
13300330119994 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 12962786549769 + 337543570225 = 3600387^2 + 580985^2 .
It is a super-2 number, since 2×133003301199942 (a number of 27 digits) contains 22 as substring.
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3325082529997 + ... + 3325082530000.
Almost surely, 213300330119994 is an apocalyptic number.
13300330119994 is a deficient number, since it is larger than the sum of its proper divisors (6650165060000).
13300330119994 is an equidigital number, since it uses as much as digits as its factorization.
13300330119994 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 6650165059999.
The product of its (nonzero) digits is 236196, while the sum is 46.
The spelling of 13300330119994 in words is "thirteen trillion, three hundred billion, three hundred thirty million, one hundred nineteen thousand, nine hundred ninety-four".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.071 sec. • engine limits •