Base | Representation |
---|---|
bin | 1011111011101100101000… |
… | …0100100110011001110011 |
3 | 1201110021120201020211112022 |
4 | 2332323022010212121303 |
5 | 3204430202400403431 |
6 | 43523202423222055 |
7 | 2522622201454226 |
oct | 276731204463163 |
9 | 51407521224468 |
10 | 13120220325491 |
11 | 41a928836a6a4 |
12 | 157a9532a792b |
13 | 7423016c53a6 |
14 | 3350447215bd |
15 | 17b448b5e07b |
hex | beeca126673 |
13120220325491 has 2 divisors, whose sum is σ = 13120220325492. Its totient is φ = 13120220325490.
The previous prime is 13120220325437. The next prime is 13120220325539. The reversal of 13120220325491 is 19452302202131.
It is a strong prime.
It is an emirp because it is prime and its reverse (19452302202131) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 13120220325491 - 218 = 13120220063347 is a prime.
It is not a weakly prime, because it can be changed into another prime (13120223325491) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6560110162745 + 6560110162746.
It is an arithmetic number, because the mean of its divisors is an integer number (6560110162746).
Almost surely, 213120220325491 is an apocalyptic number.
13120220325491 is a deficient number, since it is larger than the sum of its proper divisors (1).
13120220325491 is an equidigital number, since it uses as much as digits as its factorization.
13120220325491 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 25920, while the sum is 35.
The spelling of 13120220325491 in words is "thirteen trillion, one hundred twenty billion, two hundred twenty million, three hundred twenty-five thousand, four hundred ninety-one".
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