Base | Representation |
---|---|
bin | 11010001100100011111010… |
… | …111010001101011011010011 |
3 | 120002221012101002220100002012 |
4 | 122030203322322031123103 |
5 | 110100120010022430001 |
6 | 1045011503533213135 |
7 | 33160551660241253 |
oct | 3214437272153323 |
9 | 502835332810065 |
10 | 115212412311251 |
11 | 3378a345194022 |
12 | 10b08b2b9311ab |
13 | 4c3963a07b558 |
14 | 206444c073963 |
15 | d4be14e202bb |
hex | 68c8fae8d6d3 |
115212412311251 has 2 divisors, whose sum is σ = 115212412311252. Its totient is φ = 115212412311250.
The previous prime is 115212412311221. The next prime is 115212412311253. The reversal of 115212412311251 is 152113214212511.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 115212412311251 - 214 = 115212412294867 is a prime.
Together with 115212412311253, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (115212412311253) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57606206155625 + 57606206155626.
It is an arithmetic number, because the mean of its divisors is an integer number (57606206155626).
Almost surely, 2115212412311251 is an apocalyptic number.
115212412311251 is a deficient number, since it is larger than the sum of its proper divisors (1).
115212412311251 is an equidigital number, since it uses as much as digits as its factorization.
115212412311251 is an evil number, because the sum of its binary digits is even.
The product of its digits is 4800, while the sum is 32.
Adding to 115212412311251 its reverse (152113214212511), we get a palindrome (267325626523762).
The spelling of 115212412311251 in words is "one hundred fifteen trillion, two hundred twelve billion, four hundred twelve million, three hundred eleven thousand, two hundred fifty-one".
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