Base | Representation |
---|---|
bin | 11010001110100100000101… |
… | …011001011001000101111101 |
3 | 120010102100120201121120112222 |
4 | 122032210011121121011331 |
5 | 110104343323431224013 |
6 | 1045155023154135125 |
7 | 33203523130502021 |
oct | 3216440531310575 |
9 | 503370521546488 |
10 | 115350027211133 |
11 | 33832743212140 |
12 | 10b2b7368814a5 |
13 | 4c495cb7707b0 |
14 | 206ad84a00381 |
15 | d507bb560508 |
hex | 68e90565917d |
115350027211133 has 8 divisors (see below), whose sum is σ = 135516115884576. Its totient is φ = 96797225631600.
The previous prime is 115350027211099. The next prime is 115350027211141. The reversal of 115350027211133 is 331112720053511.
It is a happy number.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 115350027211133 - 214 = 115350027194749 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 115350027211093 and 115350027211102.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (115350027211163) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 403321773323 + ... + 403321773608.
It is an arithmetic number, because the mean of its divisors is an integer number (16939514485572).
Almost surely, 2115350027211133 is an apocalyptic number.
115350027211133 is a gapful number since it is divisible by the number (13) formed by its first and last digit.
It is an amenable number.
115350027211133 is a deficient number, since it is larger than the sum of its proper divisors (20166088673443).
115350027211133 is a wasteful number, since it uses less digits than its factorization.
115350027211133 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 806643546955.
The product of its (nonzero) digits is 18900, while the sum is 35.
Adding to 115350027211133 its reverse (331112720053511), we get a palindrome (446462747264644).
The spelling of 115350027211133 in words is "one hundred fifteen trillion, three hundred fifty billion, twenty-seven million, two hundred eleven thousand, one hundred thirty-three".
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