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115032224130041 is a prime number
BaseRepresentation
bin11010001001111100000110…
…110110110110001111111001
3120002021222021111001022202002
4122021330012312312033321
5110034141443334130131
61044353024011150345
733141542515050011
oct3211740666661771
9502258244038662
10115032224130041
113371a98a701589
1210a9a0231453b5
134c2565245a39a
142059837290c41
15d473bae446cb
hex689f06db63f9

115032224130041 has 2 divisors, whose sum is σ = 115032224130042. Its totient is φ = 115032224130040.

The previous prime is 115032224130001. The next prime is 115032224130107. The reversal of 115032224130041 is 140031422230511.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 94928075554816 + 20104148575225 = 9743104^2 + 4483765^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-115032224130041 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 115032224129986 and 115032224130013.

It is not a weakly prime, because it can be changed into another prime (115032224130001) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57516112065020 + 57516112065021.

It is an arithmetic number, because the mean of its divisors is an integer number (57516112065021).

Almost surely, 2115032224130041 is an apocalyptic number.

It is an amenable number.

115032224130041 is a deficient number, since it is larger than the sum of its proper divisors (1).

115032224130041 is an equidigital number, since it uses as much as digits as its factorization.

115032224130041 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 5760, while the sum is 29.

Adding to 115032224130041 its reverse (140031422230511), we get a palindrome (255063646360552).

The spelling of 115032224130041 in words is "one hundred fifteen trillion, thirty-two billion, two hundred twenty-four million, one hundred thirty thousand, forty-one".