Base | Representation |
---|---|
bin | 11110100111111000110… |
… | …100001010110101011111 |
3 | 21110011212012010211101001 |
4 | 132213320310022311133 |
5 | 233434314314330213 |
6 | 4250430523131131 |
7 | 305016220020505 |
oct | 36477064126537 |
9 | 7404765124331 |
10 | 2104413433183 |
11 | 741527580626 |
12 | 29ba23973aa7 |
13 | 1235a352b229 |
14 | 73bd5b24675 |
15 | 39b199061dd |
hex | 1e9f8d0ad5f |
2104413433183 has 2 divisors, whose sum is σ = 2104413433184. Its totient is φ = 2104413433182.
The previous prime is 2104413433153. The next prime is 2104413433223. The reversal of 2104413433183 is 3813343144012.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2104413433183 - 237 = 1966974479711 is a prime.
It is a super-2 number, since 2×21044134331832 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2104413433153) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1052206716591 + 1052206716592.
It is an arithmetic number, because the mean of its divisors is an integer number (1052206716592).
Almost surely, 22104413433183 is an apocalyptic number.
2104413433183 is a deficient number, since it is larger than the sum of its proper divisors (1).
2104413433183 is an equidigital number, since it uses as much as digits as its factorization.
2104413433183 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 82944, while the sum is 37.
Adding to 2104413433183 its reverse (3813343144012), we get a palindrome (5917756577195).
The spelling of 2104413433183 in words is "two trillion, one hundred four billion, four hundred thirteen million, four hundred thirty-three thousand, one hundred eighty-three".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.072 sec. • engine limits •