Search a number
-
+
131559929259433 is a prime number
BaseRepresentation
bin11101111010011100101110…
…110110011010010110101001
3122020211000011200010020102121
4131322130232312122112221
5114220434213342300213
61143445444103431241
736465614455505143
oct3572345666322651
9566730150106377
10131559929259433
1138a122863411a4
1212909242155521
13585408188b03c
14246b76ab45c93
151032297d7088d
hex77a72ed9a5a9

131559929259433 has 2 divisors, whose sum is σ = 131559929259434. Its totient is φ = 131559929259432.

The previous prime is 131559929259431. The next prime is 131559929259457. The reversal of 131559929259433 is 334952929955131.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 103385946380544 + 28173982878889 = 10167888^2 + 5307917^2 .

It is a cyclic number.

It is not a de Polignac number, because 131559929259433 - 21 = 131559929259431 is a prime.

It is a super-3 number, since 3×1315599292594333 (a number of 43 digits) contains 333 as substring.

Together with 131559929259431, it forms a pair of twin primes.

It is not a weakly prime, because it can be changed into another prime (131559929259431) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65779964629716 + 65779964629717.

It is an arithmetic number, because the mean of its divisors is an integer number (65779964629717).

Almost surely, 2131559929259433 is an apocalyptic number.

It is an amenable number.

131559929259433 is a deficient number, since it is larger than the sum of its proper divisors (1).

131559929259433 is an equidigital number, since it uses as much as digits as its factorization.

131559929259433 is an evil number, because the sum of its binary digits is even.

The product of its digits is 354294000, while the sum is 70.

The spelling of 131559929259433 in words is "one hundred thirty-one trillion, five hundred fifty-nine billion, nine hundred twenty-nine million, two hundred fifty-nine thousand, four hundred thirty-three".