Base | Representation |
---|---|
bin | 1011111100000111101101… |
… | …0111010010110100001011 |
3 | 1201110222100022222101200211 |
4 | 2333001323113102310023 |
5 | 3210040044334314043 |
6 | 43530404024341551 |
7 | 2523300301602325 |
oct | 277017327226413 |
9 | 51428308871624 |
10 | 13127489760523 |
11 | 4201378811846 |
12 | 15802419148b7 |
13 | 742bbc786c38 |
14 | 335533d76615 |
15 | 17b721e4aa9d |
hex | bf07b5d2d0b |
13127489760523 has 2 divisors, whose sum is σ = 13127489760524. Its totient is φ = 13127489760522.
The previous prime is 13127489760481. The next prime is 13127489760557. The reversal of 13127489760523 is 32506798472131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13127489760523 - 217 = 13127489629451 is a prime.
It is not a weakly prime, because it can be changed into another prime (13127489760023) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6563744880261 + 6563744880262.
It is an arithmetic number, because the mean of its divisors is an integer number (6563744880262).
It is a 1-persistent number, because it is pandigital, but 2⋅13127489760523 = 26254979521046 is not.
Almost surely, 213127489760523 is an apocalyptic number.
13127489760523 is a deficient number, since it is larger than the sum of its proper divisors (1).
13127489760523 is an equidigital number, since it uses as much as digits as its factorization.
13127489760523 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 15240960, while the sum is 58.
The spelling of 13127489760523 in words is "thirteen trillion, one hundred twenty-seven billion, four hundred eighty-nine million, seven hundred sixty thousand, five hundred twenty-three".
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