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115960035072017 is a prime number
BaseRepresentation
bin11010010111011100001100…
…101100101101100000010001
3120012120200010020000122021102
4122113130030230231200101
5110144342122434301032
61050343145451434145
733265556524662023
oct3227341454554021
9505520106018242
10115960035072017
1133a484137a9744
12110099b9155955
134c91ca490b476
14208c6d2092013
15d615bece0362
hex69770cb2d811

115960035072017 has 2 divisors, whose sum is σ = 115960035072018. Its totient is φ = 115960035072016.

The previous prime is 115960035071921. The next prime is 115960035072019. The reversal of 115960035072017 is 710270530069511.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 79475245882321 + 36484789189696 = 8914889^2 + 6040264^2 .

It is a cyclic number.

It is not a de Polignac number, because 115960035072017 - 236 = 115891315595281 is a prime.

Together with 115960035072019, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (115960035072019) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57980017536008 + 57980017536009.

It is an arithmetic number, because the mean of its divisors is an integer number (57980017536009).

Almost surely, 2115960035072017 is an apocalyptic number.

It is an amenable number.

115960035072017 is a deficient number, since it is larger than the sum of its proper divisors (1).

115960035072017 is an equidigital number, since it uses as much as digits as its factorization.

115960035072017 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 396900, while the sum is 47.

The spelling of 115960035072017 in words is "one hundred fifteen trillion, nine hundred sixty billion, thirty-five million, seventy-two thousand, seventeen".