Base | Representation |
---|---|
bin | 11010001011011001000010… |
… | …001110011011001100110111 |
3 | 120002122111211100020221121101 |
4 | 122023121002032123030313 |
5 | 110042310321112224023 |
6 | 1044510525045030531 |
7 | 33152004253344514 |
oct | 3213310216331467 |
9 | 502574740227541 |
10 | 115132004414263 |
11 | 337592330089a1 |
12 | 10ab542b3b5447 |
13 | 4c31b9459b895 |
14 | 20605c107a30b |
15 | d49caac2d1ad |
hex | 68b64239b337 |
115132004414263 has 2 divisors, whose sum is σ = 115132004414264. Its totient is φ = 115132004414262.
The previous prime is 115132004414237. The next prime is 115132004414279. The reversal of 115132004414263 is 362414400231511.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 115132004414263 - 213 = 115132004406071 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (115132004414563) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57566002207131 + 57566002207132.
It is an arithmetic number, because the mean of its divisors is an integer number (57566002207132).
Almost surely, 2115132004414263 is an apocalyptic number.
115132004414263 is a deficient number, since it is larger than the sum of its proper divisors (1).
115132004414263 is an equidigital number, since it uses as much as digits as its factorization.
115132004414263 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 69120, while the sum is 37.
Adding to 115132004414263 its reverse (362414400231511), we get a palindrome (477546404645774).
The spelling of 115132004414263 in words is "one hundred fifteen trillion, one hundred thirty-two billion, four million, four hundred fourteen thousand, two hundred sixty-three".
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