Base | Representation |
---|---|
bin | 1100010000111100110100… |
… | …0110101000100010101011 |
3 | 1202202012000202202100021012 |
4 | 3010033031012220202223 |
5 | 3231420440403112342 |
6 | 44403025314134135 |
7 | 2561166244462004 |
oct | 304171506504253 |
9 | 52665022670235 |
10 | 13485343410347 |
11 | 432a11387a486 |
12 | 161967228134b |
13 | 76a87c476b23 |
14 | 3489a08981ab |
15 | 185bb86dbb82 |
hex | c43cd1a88ab |
13485343410347 has 2 divisors, whose sum is σ = 13485343410348. Its totient is φ = 13485343410346.
The previous prime is 13485343410317. The next prime is 13485343410361. The reversal of 13485343410347 is 74301434358431.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13485343410347 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13485343410295 and 13485343410304.
It is not a weakly prime, because it can be changed into another prime (13485343410317) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6742671705173 + 6742671705174.
It is an arithmetic number, because the mean of its divisors is an integer number (6742671705174).
Almost surely, 213485343410347 is an apocalyptic number.
13485343410347 is a deficient number, since it is larger than the sum of its proper divisors (1).
13485343410347 is an equidigital number, since it uses as much as digits as its factorization.
13485343410347 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5806080, while the sum is 50.
Adding to 13485343410347 its reverse (74301434358431), we get a palindrome (87786777768778).
The spelling of 13485343410347 in words is "thirteen trillion, four hundred eighty-five billion, three hundred forty-three million, four hundred ten thousand, three hundred forty-seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.065 sec. • engine limits •