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3331122223147 is a prime number
BaseRepresentation
bin110000011110010110010…
…110110101100000101011
3102210110012122011001022121
4300132112112311200223
5414034114242120042
611030144005532111
7462444212500045
oct60362626654053
912713178131277
103331122223147
1110747a271a954
12459716114037
131b217b385864
14b7327528a95
155b9b4302367
hex307965b582b

3331122223147 has 2 divisors, whose sum is σ = 3331122223148. Its totient is φ = 3331122223146.

The previous prime is 3331122223129. The next prime is 3331122223199. The reversal of 3331122223147 is 7413222211333.

3331122223147 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is an emirp because it is prime and its reverse (7413222211333) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 3331122223147 - 27 = 3331122223019 is a prime.

It is a super-3 number, since 3×33311222231473 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (3331122223747) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1665561111573 + 1665561111574.

It is an arithmetic number, because the mean of its divisors is an integer number (1665561111574).

Almost surely, 23331122223147 is an apocalyptic number.

3331122223147 is a deficient number, since it is larger than the sum of its proper divisors (1).

3331122223147 is an equidigital number, since it uses as much as digits as its factorization.

3331122223147 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 36288, while the sum is 34.

The spelling of 3331122223147 in words is "three trillion, three hundred thirty-one billion, one hundred twenty-two million, two hundred twenty-three thousand, one hundred forty-seven".