Base | Representation |
---|---|
bin | 1011111011111011111100… |
… | …0011101100010011011111 |
3 | 1201110200012011021121110021 |
4 | 2332332333003230103133 |
5 | 3210012112102133224 |
6 | 43525122330453011 |
7 | 2523126100432156 |
oct | 276767703542337 |
9 | 51420164247407 |
10 | 13124330505439 |
11 | 41aaaa746872a |
12 | 157b6bb894167 |
13 | 7428080b302a |
14 | 33531455659d |
15 | 17b5d98ea1e4 |
hex | befbf0ec4df |
13124330505439 has 2 divisors, whose sum is σ = 13124330505440. Its totient is φ = 13124330505438.
The previous prime is 13124330505379. The next prime is 13124330505467. The reversal of 13124330505439 is 93450503342131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13124330505439 - 219 = 13124329981151 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13124330505395 and 13124330505404.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13124330535439) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6562165252719 + 6562165252720.
It is an arithmetic number, because the mean of its divisors is an integer number (6562165252720).
Almost surely, 213124330505439 is an apocalyptic number.
13124330505439 is a deficient number, since it is larger than the sum of its proper divisors (1).
13124330505439 is an equidigital number, since it uses as much as digits as its factorization.
13124330505439 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 583200, while the sum is 43.
The spelling of 13124330505439 in words is "thirteen trillion, one hundred twenty-four billion, three hundred thirty million, five hundred five thousand, four hundred thirty-nine".
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