Base | Representation |
---|---|
bin | 1011000110111001011101… |
… | …0001101011000010000011 |
3 | 1121020120020000210122121012 |
4 | 2301232113101223002003 |
5 | 3100044442240433401 |
6 | 41550345013551135 |
7 | 2400240514156133 |
oct | 261562721530203 |
9 | 47216200718535 |
10 | 12213130014851 |
11 | 3989617552547 |
12 | 1452ba06454ab |
13 | 6a7902002bcc |
14 | 3031937d0ac3 |
15 | 162a58e298bb |
hex | b1b9746b083 |
12213130014851 has 2 divisors, whose sum is σ = 12213130014852. Its totient is φ = 12213130014850.
The previous prime is 12213130014847. The next prime is 12213130014859. The reversal of 12213130014851 is 15841003131221.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 12213130014851 - 22 = 12213130014847 is a prime.
It is not a weakly prime, because it can be changed into another prime (12213130014859) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6106565007425 + 6106565007426.
It is an arithmetic number, because the mean of its divisors is an integer number (6106565007426).
Almost surely, 212213130014851 is an apocalyptic number.
12213130014851 is a deficient number, since it is larger than the sum of its proper divisors (1).
12213130014851 is an equidigital number, since it uses as much as digits as its factorization.
12213130014851 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5760, while the sum is 32.
The spelling of 12213130014851 in words is "twelve trillion, two hundred thirteen billion, one hundred thirty million, fourteen thousand, eight hundred fifty-one".
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