Base | Representation |
---|---|
bin | 111010011000100101001… |
… | …011110111111100100001 |
3 | 112012112222202212101021022 |
4 | 322120211023313330201 |
5 | 1011213312032333233 |
6 | 12311051034535225 |
7 | 562603104364445 |
oct | 72304513677441 |
9 | 15175882771238 |
10 | 4012123324193 |
11 | 1307594025566 |
12 | 5496abb97515 |
13 | 23145967c24a |
14 | dc28b4a0825 |
15 | 6e57048b398 |
hex | 3a6252f7f21 |
4012123324193 has 2 divisors, whose sum is σ = 4012123324194. Its totient is φ = 4012123324192.
The previous prime is 4012123324151. The next prime is 4012123324211. The reversal of 4012123324193 is 3914233212104.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 2455887034129 + 1556236290064 = 1567127^2 + 1247492^2 .
It is a cyclic number.
It is not a de Polignac number, because 4012123324193 - 212 = 4012123320097 is a prime.
It is not a weakly prime, because it can be changed into another prime (4012123324393) 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, 2006061662096 + 2006061662097.
It is an arithmetic number, because the mean of its divisors is an integer number (2006061662097).
Almost surely, 24012123324193 is an apocalyptic number.
It is an amenable number.
4012123324193 is a deficient number, since it is larger than the sum of its proper divisors (1).
4012123324193 is an equidigital number, since it uses as much as digits as its factorization.
4012123324193 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 31104, while the sum is 35.
Adding to 4012123324193 its reverse (3914233212104), we get a palindrome (7926356536297).
The spelling of 4012123324193 in words is "four trillion, twelve billion, one hundred twenty-three million, three hundred twenty-four thousand, one hundred ninety-three".
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