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3433113231137 is a prime number
BaseRepresentation
bin110001111101010101011…
…111101001111100100001
3110011012110120020002001222
4301331111133221330201
5422222003441344022
611145052255104425
7503014513656422
oct61752537517441
913135416202058
103433113231137
111103a7aa107a7
1247543a812115
131bb983511825
14bc240b83649
155e483225242
hex31f557e9f21

3433113231137 has 2 divisors, whose sum is σ = 3433113231138. Its totient is φ = 3433113231136.

The previous prime is 3433113231103. The next prime is 3433113231139. The reversal of 3433113231137 is 7311323113343.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 2117283997921 + 1315829233216 = 1455089^2 + 1147096^2 .

It is an emirp because it is prime and its reverse (7311323113343) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 3433113231137 - 212 = 3433113227041 is a prime.

Together with 3433113231139, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 3433113231097 and 3433113231106.

It is not a weakly prime, because it can be changed into another prime (3433113231139) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1716556615568 + 1716556615569.

It is an arithmetic number, because the mean of its divisors is an integer number (1716556615569).

Almost surely, 23433113231137 is an apocalyptic number.

It is an amenable number.

3433113231137 is a deficient number, since it is larger than the sum of its proper divisors (1).

3433113231137 is an equidigital number, since it uses as much as digits as its factorization.

3433113231137 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 40824, while the sum is 35.

The spelling of 3433113231137 in words is "three trillion, four hundred thirty-three billion, one hundred thirteen million, two hundred thirty-one thousand, one hundred thirty-seven".