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3440851327049 is a prime number
BaseRepresentation
bin110010000100100010101…
…110001001100001001001
3110011221102211211002002212
4302010202232021201021
5422333330414431144
611152412201234505
7503410334465003
oct62044256114111
913157384732085
103440851327049
111107290964794
12476a3a192a35
131bc617704ac6
14bc776753373
155e78773949e
hex32122b89849

3440851327049 has 2 divisors, whose sum is σ = 3440851327050. Its totient is φ = 3440851327048.

The previous prime is 3440851327019. The next prime is 3440851327051. The reversal of 3440851327049 is 9407231580443.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 2478798336400 + 962052990649 = 1574420^2 + 980843^2 .

It is an emirp because it is prime and its reverse (9407231580443) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 3440851327049 - 236 = 3372131850313 is a prime.

Together with 3440851327051, 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 = 3440851326988 and 3440851327006.

It is not a weakly prime, because it can be changed into another prime (3440851327009) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1720425663524 + 1720425663525.

It is an arithmetic number, because the mean of its divisors is an integer number (1720425663525).

Almost surely, 23440851327049 is an apocalyptic number.

It is an amenable number.

3440851327049 is a deficient number, since it is larger than the sum of its proper divisors (1).

3440851327049 is an equidigital number, since it uses as much as digits as its factorization.

3440851327049 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2903040, while the sum is 50.

The spelling of 3440851327049 in words is "three trillion, four hundred forty billion, eight hundred fifty-one million, three hundred twenty-seven thousand, forty-nine".