Base | Representation |
---|---|
bin | 10011100000011000000… |
… | …100101101111110101001 |
3 | 11202010220010110112200102 |
4 | 103200120010231332221 |
5 | 133430202010431231 |
6 | 2503441533515145 |
7 | 165562124225213 |
oct | 23403004557651 |
9 | 4663803415612 |
10 | 1340433686441 |
11 | 477524964283 |
12 | 197950642ab5 |
13 | 9952cc841a5 |
14 | 48c3d5a95b3 |
15 | 24d03a26acb |
hex | 1381812dfa9 |
1340433686441 has 2 divisors, whose sum is σ = 1340433686442. Its totient is φ = 1340433686440.
The previous prime is 1340433686437. The next prime is 1340433686621. The reversal of 1340433686441 is 1446863340431.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 1235843539225 + 104590147216 = 1111685^2 + 323404^2 .
It is a cyclic number.
It is not a de Polignac number, because 1340433686441 - 22 = 1340433686437 is a prime.
It is not a weakly prime, because it can be changed into another prime (1340433686401) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 670216843220 + 670216843221.
It is an arithmetic number, because the mean of its divisors is an integer number (670216843221).
Almost surely, 21340433686441 is an apocalyptic number.
It is an amenable number.
1340433686441 is a deficient number, since it is larger than the sum of its proper divisors (1).
1340433686441 is an equidigital number, since it uses as much as digits as its factorization.
1340433686441 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1990656, while the sum is 47.
The spelling of 1340433686441 in words is "one trillion, three hundred forty billion, four hundred thirty-three million, six hundred eighty-six thousand, four hundred forty-one".
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