Search a number
-
+
3331122223129 is a prime number
BaseRepresentation
bin110000011110010110010…
…110110101100000011001
3102210110012122011001021221
4300132112112311200121
5414034114242120004
611030144005532041
7462444212500021
oct60362626654031
912713178131257
103331122223129
1110747a271a938
12459716114021
131b217b38584c
14b7327528a81
155b9b4302354
hex307965b5819

3331122223129 has 2 divisors, whose sum is σ = 3331122223130. Its totient is φ = 3331122223128.

The previous prime is 3331122223109. The next prime is 3331122223147. The reversal of 3331122223129 is 9213222211333.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 2611543264729 + 719578958400 = 1616027^2 + 848280^2 .

It is a cyclic number.

It is not a de Polignac number, because 3331122223129 - 27 = 3331122223001 is a prime.

It is a super-4 number, since 4×33311222231294 (a number of 51 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.

It is a junction number, because it is equal to n+sod(n) for n = 3331122223094 and 3331122223103.

It is not a weakly prime, because it can be changed into another prime (3331122223109) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1665561111564 + 1665561111565.

It is an arithmetic number, because the mean of its divisors is an integer number (1665561111565).

Almost surely, 23331122223129 is an apocalyptic number.

It is an amenable number.

3331122223129 is a deficient number, since it is larger than the sum of its proper divisors (1).

3331122223129 is an equidigital number, since it uses as much as digits as its factorization.

3331122223129 is an evil number, because the sum of its binary digits is even.

The product of its digits is 23328, while the sum is 34.

The spelling of 3331122223129 in words is "three trillion, three hundred thirty-one billion, one hundred twenty-two million, two hundred twenty-three thousand, one hundred twenty-nine".