Base | Representation |
---|---|
bin | 1011111011000110111110… |
… | …0100110001111111111011 |
3 | 1201102022111020211210102021 |
4 | 2332301233210301333323 |
5 | 3204244002144202342 |
6 | 43514411413015311 |
7 | 2522113541122123 |
oct | 276615744617773 |
9 | 51368436753367 |
10 | 13110112100347 |
11 | 41a4a70556519 |
12 | 15789b2024537 |
13 | 741380474692 |
14 | 334766069c83 |
15 | 17b056524567 |
hex | bec6f931ffb |
13110112100347 has 2 divisors, whose sum is σ = 13110112100348. Its totient is φ = 13110112100346.
The previous prime is 13110112100329. The next prime is 13110112100389. The reversal of 13110112100347 is 74300121101131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13110112100347 - 27 = 13110112100219 is a prime.
It is not a weakly prime, because it can be changed into another prime (13110112100747) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6555056050173 + 6555056050174.
It is an arithmetic number, because the mean of its divisors is an integer number (6555056050174).
Almost surely, 213110112100347 is an apocalyptic number.
13110112100347 is a deficient number, since it is larger than the sum of its proper divisors (1).
13110112100347 is an equidigital number, since it uses as much as digits as its factorization.
13110112100347 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 504, while the sum is 25.
Adding to 13110112100347 its reverse (74300121101131), we get a palindrome (87410233201478).
The spelling of 13110112100347 in words is "thirteen trillion, one hundred ten billion, one hundred twelve million, one hundred thousand, three hundred forty-seven".
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