Base | Representation |
---|---|
bin | 10001011010001011100… |
… | …100011100011010110111 |
3 | 11020100222011110121012201 |
4 | 101122023210130122313 |
5 | 124100102134314103 |
6 | 2313331523325331 |
7 | 152301324264613 |
oct | 21321344343267 |
9 | 4210864417181 |
10 | 1196342494903 |
11 | 421403171406 |
12 | 173a38820847 |
13 | 88a789223ca |
14 | 41c9088a343 |
15 | 211bda6a51d |
hex | 1168b91c6b7 |
1196342494903 has 2 divisors, whose sum is σ = 1196342494904. Its totient is φ = 1196342494902.
The previous prime is 1196342494883. The next prime is 1196342494913. The reversal of 1196342494903 is 3094942436911.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1196342494903 - 221 = 1196340397751 is a prime.
It is a super-3 number, since 3×11963424949033 (a number of 37 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1196342494913) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 598171247451 + 598171247452.
It is an arithmetic number, because the mean of its divisors is an integer number (598171247452).
Almost surely, 21196342494903 is an apocalyptic number.
1196342494903 is a deficient number, since it is larger than the sum of its proper divisors (1).
1196342494903 is an equidigital number, since it uses as much as digits as its factorization.
1196342494903 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5038848, while the sum is 55.
The spelling of 1196342494903 in words is "one trillion, one hundred ninety-six billion, three hundred forty-two million, four hundred ninety-four thousand, nine hundred three".
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