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113242304012137 is a prime number
BaseRepresentation
bin11001101111111001000111…
…010011011101111101101001
3112211221212011212120020000221
4121233321013103131331221
5104320330214311342022
61040502443555130041
732565325601063206
oct3157710723357551
9484855155506027
10113242304012137
113309a878437573
121084b14a5a3921
134b2591b621171
141dd6d574bacad
15d15a5b2c7cc7
hex66fe474ddf69

113242304012137 has 2 divisors, whose sum is σ = 113242304012138. Its totient is φ = 113242304012136.

The previous prime is 113242304012039. The next prime is 113242304012147. The reversal of 113242304012137 is 731210403242311.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 59215041107161 + 54027262904976 = 7695131^2 + 7350324^2 .

It is an emirp because it is prime and its reverse (731210403242311) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 113242304012137 - 27 = 113242304012009 is a prime.

It is not a weakly prime, because it can be changed into another prime (113242304012147) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56621152006068 + 56621152006069.

It is an arithmetic number, because the mean of its divisors is an integer number (56621152006069).

Almost surely, 2113242304012137 is an apocalyptic number.

It is an amenable number.

113242304012137 is a deficient number, since it is larger than the sum of its proper divisors (1).

113242304012137 is an equidigital number, since it uses as much as digits as its factorization.

113242304012137 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 24192, while the sum is 34.

Adding to 113242304012137 its reverse (731210403242311), we get a palindrome (844452707254448).

The spelling of 113242304012137 in words is "one hundred thirteen trillion, two hundred forty-two billion, three hundred four million, twelve thousand, one hundred thirty-seven".