Base | Representation |
---|---|
bin | 1011001101110010001000… |
… | …0101110000111100010111 |
3 | 1121122212120002000010101222 |
4 | 2303130202011300330113 |
5 | 3104014223244341401 |
6 | 42120551052040555 |
7 | 2411626105262216 |
oct | 263344205607427 |
9 | 47585502003358 |
10 | 12331423043351 |
11 | 3a247aa912715 |
12 | 1471ab476315b |
13 | 6b5b0466c769 |
14 | 308bb615907d |
15 | 165b7e100c1b |
hex | b3722170f17 |
12331423043351 has 2 divisors, whose sum is σ = 12331423043352. Its totient is φ = 12331423043350.
The previous prime is 12331423043347. The next prime is 12331423043477. The reversal of 12331423043351 is 15334032413321.
12331423043351 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 12331423043351 - 22 = 12331423043347 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12331423043551) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6165711521675 + 6165711521676.
It is an arithmetic number, because the mean of its divisors is an integer number (6165711521676).
Almost surely, 212331423043351 is an apocalyptic number.
12331423043351 is a deficient number, since it is larger than the sum of its proper divisors (1).
12331423043351 is an equidigital number, since it uses as much as digits as its factorization.
12331423043351 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 77760, while the sum is 35.
Adding to 12331423043351 its reverse (15334032413321), we get a palindrome (27665455456672).
The spelling of 12331423043351 in words is "twelve trillion, three hundred thirty-one billion, four hundred twenty-three million, forty-three thousand, three hundred fifty-one".
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