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514442535240113 is a prime number
BaseRepresentation
bin111010011111000011111110…
…0100100100000000110110001
32111110111020002210122001022022
41310332013330210200012301
51014412111303010140423
65022043345454521225
7213234150412110614
oct16476077444400661
92443436083561268
10514442535240113
11139a0a943721257
12498464b0206215
1319108952325aac
149107219c4017b
153e71c6e568dc8
hex1d3e1fc9201b1

514442535240113 has 2 divisors, whose sum is σ = 514442535240114. Its totient is φ = 514442535240112.

The previous prime is 514442535240059. The next prime is 514442535240173. The reversal of 514442535240113 is 311042535244415.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 478269912800689 + 36172622439424 = 21869383^2 + 6014368^2 .

It is a cyclic number.

It is not a de Polignac number, because 514442535240113 - 210 = 514442535239089 is a prime.

It is a super-2 number, since 2×5144425352401132 (a number of 30 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (514442535240173) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 257221267620056 + 257221267620057.

It is an arithmetic number, because the mean of its divisors is an integer number (257221267620057).

Almost surely, 2514442535240113 is an apocalyptic number.

It is an amenable number.

514442535240113 is a deficient number, since it is larger than the sum of its proper divisors (1).

514442535240113 is an equidigital number, since it uses as much as digits as its factorization.

514442535240113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1152000, while the sum is 44.

The spelling of 514442535240113 in words is "five hundred fourteen trillion, four hundred forty-two billion, five hundred thirty-five million, two hundred forty thousand, one hundred thirteen".