Base | Representation |
---|---|
bin | 11110100001110100000100… |
… | …011101110111011101000111 |
3 | 122121101121120000111000212012 |
4 | 132201310010131313131013 |
5 | 120044244304140002224 |
6 | 1153320302241244435 |
7 | 40165221021024461 |
oct | 3641640435673507 |
9 | 577347500430765 |
10 | 134265047578439 |
11 | 3986553a091488 |
12 | 1308557093371b |
13 | 59bc1a7aa696b |
14 | 252266a59b131 |
15 | 107c81e42e80e |
hex | 7a1d04777747 |
134265047578439 has 2 divisors, whose sum is σ = 134265047578440. Its totient is φ = 134265047578438.
The previous prime is 134265047578429. The next prime is 134265047578507. The reversal of 134265047578439 is 934875740562431.
It is an a-pointer prime, because the next prime (134265047578507) can be obtained adding 134265047578439 to its sum of digits (68).
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 134265047578439 - 240 = 133165535950663 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (134265047578429) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67132523789219 + 67132523789220.
It is an arithmetic number, because the mean of its divisors is an integer number (67132523789220).
It is a 1-persistent number, because it is pandigital, but 2⋅134265047578439 = 268530095156878 is not.
Almost surely, 2134265047578439 is an apocalyptic number.
134265047578439 is a deficient number, since it is larger than the sum of its proper divisors (1).
134265047578439 is an equidigital number, since it uses as much as digits as its factorization.
134265047578439 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 609638400, while the sum is 68.
The spelling of 134265047578439 in words is "one hundred thirty-four trillion, two hundred sixty-five billion, forty-seven million, five hundred seventy-eight thousand, four hundred thirty-nine".
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