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1179665594113 is a prime number
BaseRepresentation
bin10001001010101001100…
…011000111011100000001
311011202220220221021002111
4101022221203013130001
5123311423343002423
62301533023415321
7151141133050606
oct21125143073401
94152826837074
101179665594113
11415325523a76
12170763754541
138731b856a86
144114ba746ad
1520a4491660d
hex112a98c7701

1179665594113 has 2 divisors, whose sum is σ = 1179665594114. Its totient is φ = 1179665594112.

The previous prime is 1179665594087. The next prime is 1179665594137. The reversal of 1179665594113 is 3114955669711.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 950404662769 + 229260931344 = 974887^2 + 478812^2 .

It is a cyclic number.

It is not a de Polignac number, because 1179665594113 - 233 = 1171075659521 is a prime.

It is a super-3 number, since 3×11796655941133 (a number of 37 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 (1179665594713) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 589832797056 + 589832797057.

It is an arithmetic number, because the mean of its divisors is an integer number (589832797057).

Almost surely, 21179665594113 is an apocalyptic number.

It is an amenable number.

1179665594113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1179665594113 is an equidigital number, since it uses as much as digits as its factorization.

1179665594113 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 6123600, while the sum is 58.

The spelling of 1179665594113 in words is "one trillion, one hundred seventy-nine billion, six hundred sixty-five million, five hundred ninety-four thousand, one hundred thirteen".