Base | Representation |
---|---|
bin | 1010000001101111110100… |
… | …1000101101011101011111 |
3 | 1110000222210110211101120101 |
4 | 2200123331020231131133 |
5 | 2421113433022004133 |
6 | 35240513131001531 |
7 | 2215353034153354 |
oct | 240337510553537 |
9 | 43028713741511 |
10 | 11025133000543 |
11 | 3570807177242 |
12 | 12a08b28992a7 |
13 | 61c8863869cc |
14 | 2a189546062b |
15 | 141bc7d9e37d |
hex | a06fd22d75f |
11025133000543 has 2 divisors, whose sum is σ = 11025133000544. Its totient is φ = 11025133000542.
The previous prime is 11025133000517. The next prime is 11025133000547. The reversal of 11025133000543 is 34500033152011.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 11025133000543 - 213 = 11025132992351 is a prime.
It is a super-3 number, since 3×110251330005433 (a number of 40 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11025133000547) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5512566500271 + 5512566500272.
It is an arithmetic number, because the mean of its divisors is an integer number (5512566500272).
Almost surely, 211025133000543 is an apocalyptic number.
11025133000543 is a deficient number, since it is larger than the sum of its proper divisors (1).
11025133000543 is an equidigital number, since it uses as much as digits as its factorization.
11025133000543 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5400, while the sum is 28.
Adding to 11025133000543 its reverse (34500033152011), we get a palindrome (45525166152554).
The spelling of 11025133000543 in words is "eleven trillion, twenty-five billion, one hundred thirty-three million, five hundred forty-three", and thus it is an aban number.
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