Base | Representation |
---|---|
bin | 100110110010000001100… |
… | …0010111011010111100001 |
3 | 200212112221012212200202122 |
4 | 1031210003002323113201 |
5 | 1144312024002412231 |
6 | 15200341111154025 |
7 | 1060041661220642 |
oct | 115440302732741 |
9 | 20775835780678 |
10 | 5330105513441 |
11 | 1775537674685 |
12 | 721015675915 |
13 | 2c881c0ac147 |
14 | 145d9ad5b6c9 |
15 | 939acdec77b |
hex | 4d9030bb5e1 |
5330105513441 has 2 divisors, whose sum is σ = 5330105513442. Its totient is φ = 5330105513440.
The previous prime is 5330105513423. The next prime is 5330105513443. The reversal of 5330105513441 is 1443155010335.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4543735612816 + 786369900625 = 2131604^2 + 886775^2 .
It is a cyclic number.
It is not a de Polignac number, because 5330105513441 - 242 = 932059002337 is a prime.
Together with 5330105513443, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (5330105513443) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2665052756720 + 2665052756721.
It is an arithmetic number, because the mean of its divisors is an integer number (2665052756721).
Almost surely, 25330105513441 is an apocalyptic number.
It is an amenable number.
5330105513441 is a deficient number, since it is larger than the sum of its proper divisors (1).
5330105513441 is an equidigital number, since it uses as much as digits as its factorization.
5330105513441 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 54000, while the sum is 35.
The spelling of 5330105513441 in words is "five trillion, three hundred thirty billion, one hundred five million, five hundred thirteen thousand, four hundred forty-one".
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