Base | Representation |
---|---|
bin | 11110000010011101011100… |
… | …000101100001101000111011 |
3 | 122022202122010222202011101101 |
4 | 132002131130011201220323 |
5 | 114303444142131310133 |
6 | 1144550411121232231 |
7 | 36553443646621102 |
oct | 3602353405415073 |
9 | 568678128664341 |
10 | 132110444010043 |
11 | 391047a7593474 |
12 | 12997a80492677 |
13 | 5893c662c6895 |
14 | 248a272bb1039 |
15 | 104176862737d |
hex | 78275c161a3b |
132110444010043 has 2 divisors, whose sum is σ = 132110444010044. Its totient is φ = 132110444010042.
The previous prime is 132110444010011. The next prime is 132110444010107. The reversal of 132110444010043 is 340010444011231.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 132110444010043 - 25 = 132110444010011 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 132110444009993 and 132110444010020.
It is not a weakly prime, because it can be changed into another prime (132110444010443) 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, 66055222005021 + 66055222005022.
It is an arithmetic number, because the mean of its divisors is an integer number (66055222005022).
Almost surely, 2132110444010043 is an apocalyptic number.
132110444010043 is a deficient number, since it is larger than the sum of its proper divisors (1).
132110444010043 is an equidigital number, since it uses as much as digits as its factorization.
132110444010043 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 4608, while the sum is 28.
Adding to 132110444010043 its reverse (340010444011231), we get a palindrome (472120888021274).
The spelling of 132110444010043 in words is "one hundred thirty-two trillion, one hundred ten billion, four hundred forty-four million, ten thousand, forty-three".
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