Base | Representation |
---|---|
bin | 10011000111110011010… |
… | …000100111011001110111 |
3 | 11122121210010211121021212 |
4 | 103013303100213121313 |
5 | 133012131314310341 |
6 | 2443355311232035 |
7 | 163636201533422 |
oct | 23076320473167 |
9 | 4577703747255 |
10 | 1314046244471 |
11 | 4673139365a4 |
12 | 19280751861b |
13 | 96bb61781bc |
14 | 47858c5c9b9 |
15 | 242ac1405eb |
hex | 131f3427677 |
1314046244471 has 2 divisors, whose sum is σ = 1314046244472. Its totient is φ = 1314046244470.
The previous prime is 1314046244437. The next prime is 1314046244491. The reversal of 1314046244471 is 1744426404131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1314046244471 - 214 = 1314046228087 is a prime.
It is a super-3 number, since 3×13140462444713 (a number of 37 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1314046244411) 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 as a sum of consecutive naturals, namely, 657023122235 + 657023122236.
It is an arithmetic number, because the mean of its divisors is an integer number (657023122236).
Almost surely, 21314046244471 is an apocalyptic number.
1314046244471 is a deficient number, since it is larger than the sum of its proper divisors (1).
1314046244471 is an equidigital number, since it uses as much as digits as its factorization.
1314046244471 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 258048, while the sum is 41.
The spelling of 1314046244471 in words is "one trillion, three hundred fourteen billion, forty-six million, two hundred forty-four thousand, four hundred seventy-one".
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