Base | Representation |
---|---|
bin | 11010001110010011100001… |
… | …011001101111110001101011 |
3 | 120010100122200100100011110111 |
4 | 122032103201121233301223 |
5 | 110104100413311030313 |
6 | 1045142522354355151 |
7 | 33202323336200524 |
oct | 3216234131576153 |
9 | 503318610304414 |
10 | 115332243455083 |
11 | 338261477a7a96 |
12 | 10b281b2b8bab7 |
13 | 4c47a26294115 |
14 | 206a178bc384b |
15 | d500ca16e43d |
hex | 68e4e166fc6b |
115332243455083 has 2 divisors, whose sum is σ = 115332243455084. Its totient is φ = 115332243455082.
The previous prime is 115332243455081. The next prime is 115332243455137. The reversal of 115332243455083 is 380554342233511.
It is a weak prime.
It is an emirp because it is prime and its reverse (380554342233511) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 115332243455083 - 21 = 115332243455081 is a prime.
Together with 115332243455081, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (115332243455081) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57666121727541 + 57666121727542.
It is an arithmetic number, because the mean of its divisors is an integer number (57666121727542).
Almost surely, 2115332243455083 is an apocalyptic number.
115332243455083 is a deficient number, since it is larger than the sum of its proper divisors (1).
115332243455083 is an equidigital number, since it uses as much as digits as its factorization.
115332243455083 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5184000, while the sum is 49.
Adding to 115332243455083 its reverse (380554342233511), we get a palindrome (495886585688594).
The spelling of 115332243455083 in words is "one hundred fifteen trillion, three hundred thirty-two billion, two hundred forty-three million, four hundred fifty-five thousand, eighty-three".
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