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117110750593 is a prime number
BaseRepresentation
bin110110100010001011…
…0000011110110000001
3102012021122112112102011
41231010112003312001
53404320323004333
6125444441452521
711314053645514
oct1550426036601
9365248475364
10117110750593
1145736a692a1
121a8441a1141
13b07473ba35
14594d72cd7b
1530a64db9cd
hex1b44583d81

117110750593 has 2 divisors, whose sum is σ = 117110750594. Its totient is φ = 117110750592.

The previous prime is 117110750587. The next prime is 117110750599. The reversal of 117110750593 is 395057011711.

It is a balanced prime because it is at equal distance from previous prime (117110750587) and next prime (117110750599).

It can be written as a sum of positive squares in only one way, i.e., 117054305424 + 56445169 = 342132^2 + 7513^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-117110750593 is a prime.

It is a super-3 number, since 3×1171107505933 (a number of 34 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (117110750599) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 58555375296 + 58555375297.

It is an arithmetic number, because the mean of its divisors is an integer number (58555375297).

Almost surely, 2117110750593 is an apocalyptic number.

It is an amenable number.

117110750593 is a deficient number, since it is larger than the sum of its proper divisors (1).

117110750593 is an equidigital number, since it uses as much as digits as its factorization.

117110750593 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 33075, while the sum is 40.

The spelling of 117110750593 in words is "one hundred seventeen billion, one hundred ten million, seven hundred fifty thousand, five hundred ninety-three".