Base | Representation |
---|---|
bin | 1100011010000001000110… |
… | …1110110000000101011011 |
3 | 1210022002002122211212202111 |
4 | 3012200101232300011123 |
5 | 3241444000040202342 |
6 | 45002350250013151 |
7 | 2605352340316432 |
oct | 306402156600533 |
9 | 53262078755674 |
10 | 13641113600347 |
11 | 438a1890057a2 |
12 | 16438a54171b7 |
13 | 77c4752cb39c |
14 | 35233a716119 |
15 | 189c83b5c217 |
hex | c6811bb015b |
13641113600347 has 2 divisors, whose sum is σ = 13641113600348. Its totient is φ = 13641113600346.
The previous prime is 13641113600297. The next prime is 13641113600419. The reversal of 13641113600347 is 74300631114631.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13641113600347 is a prime.
It is not a weakly prime, because it can be changed into another prime (13641113604347) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6820556800173 + 6820556800174.
It is an arithmetic number, because the mean of its divisors is an integer number (6820556800174).
Almost surely, 213641113600347 is an apocalyptic number.
13641113600347 is a deficient number, since it is larger than the sum of its proper divisors (1).
13641113600347 is an equidigital number, since it uses as much as digits as its factorization.
13641113600347 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 108864, while the sum is 40.
Adding to 13641113600347 its reverse (74300631114631), we get a palindrome (87941744714978).
The spelling of 13641113600347 in words is "thirteen trillion, six hundred forty-one billion, one hundred thirteen million, six hundred thousand, three hundred forty-seven".
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