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114012533541113 is a prime number
BaseRepresentation
bin11001111011000110011100…
…100011111100100011111001
3112221200111021111022100012122
4121323012130203330203321
5104420440132041303423
61042252345043051025
733005063603051114
oct3173063443744371
9487614244270178
10114012533541113
11333674979797a9
1210954486970a75
134b8044992b4bc
142022343aa557b
15d2aadabce8c8
hex67b19c8fc8f9

114012533541113 has 2 divisors, whose sum is σ = 114012533541114. Its totient is φ = 114012533541112.

The previous prime is 114012533541073. The next prime is 114012533541149. The reversal of 114012533541113 is 311145335210411.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 100457081117584 + 13555452423529 = 10022828^2 + 3681773^2 .

It is a cyclic number.

It is not a de Polignac number, because 114012533541113 - 210 = 114012533540089 is a prime.

It is a super-2 number, since 2×1140125335411132 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (114012533546113) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57006266770556 + 57006266770557.

It is an arithmetic number, because the mean of its divisors is an integer number (57006266770557).

Almost surely, 2114012533541113 is an apocalyptic number.

It is an amenable number.

114012533541113 is a deficient number, since it is larger than the sum of its proper divisors (1).

114012533541113 is an equidigital number, since it uses as much as digits as its factorization.

114012533541113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 21600, while the sum is 35.

Adding to 114012533541113 its reverse (311145335210411), we get a palindrome (425157868751524).

The spelling of 114012533541113 in words is "one hundred fourteen trillion, twelve billion, five hundred thirty-three million, five hundred forty-one thousand, one hundred thirteen".