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11314344041173 is a prime number
BaseRepresentation
bin1010010010100101001101…
…1101010001101011010101
31111001122022002102111101201
42210221103131101223111
52440333234033304143
640021423215152501
72245301664353005
oct244512335215325
944048262374351
1011314344041173
113672428279698
121328966b15731
13640c26c0c694
142b188d74a005
151494a32a964d
hexa4a53751ad5

11314344041173 has 2 divisors, whose sum is σ = 11314344041174. Its totient is φ = 11314344041172.

The previous prime is 11314344041159. The next prime is 11314344041219. The reversal of 11314344041173 is 37114044341311.

11314344041173 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 can be written as a sum of positive squares in only one way, i.e., 8138599069329 + 3175744971844 = 2852823^2 + 1782062^2 .

It is a cyclic number.

It is not a de Polignac number, because 11314344041173 - 217 = 11314343910101 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11314344044173) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5657172020586 + 5657172020587.

It is an arithmetic number, because the mean of its divisors is an integer number (5657172020587).

Almost surely, 211314344041173 is an apocalyptic number.

It is an amenable number.

11314344041173 is a deficient number, since it is larger than the sum of its proper divisors (1).

11314344041173 is an equidigital number, since it uses as much as digits as its factorization.

11314344041173 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 48384, while the sum is 37.

Adding to 11314344041173 its reverse (37114044341311), we get a palindrome (48428388382484).

The spelling of 11314344041173 in words is "eleven trillion, three hundred fourteen billion, three hundred forty-four million, forty-one thousand, one hundred seventy-three".