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132355233524147 is a prime number
BaseRepresentation
bin11110000110000001011010…
…101011011010100110110011
3122100122000222210101220120112
4132012001122223122212303
5114322002004240233042
61145255053242131535
736610231033303634
oct3606013253324663
9570560883356515
10132355233524147
11391995a2915527
1212a173b7ab0bab
1358b1077a3ab0a
1424980555ad88b
151047ce3cad982
hex78605aada9b3

132355233524147 has 2 divisors, whose sum is σ = 132355233524148. Its totient is φ = 132355233524146.

The previous prime is 132355233524141. The next prime is 132355233524149. The reversal of 132355233524147 is 741425332553231.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 132355233524147 - 26 = 132355233524083 is a prime.

Together with 132355233524149, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 132355233524095 and 132355233524104.

It is not a weakly prime, because it can be changed into another prime (132355233524141) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66177616762073 + 66177616762074.

It is an arithmetic number, because the mean of its divisors is an integer number (66177616762074).

Almost surely, 2132355233524147 is an apocalyptic number.

132355233524147 is a deficient number, since it is larger than the sum of its proper divisors (1).

132355233524147 is an equidigital number, since it uses as much as digits as its factorization.

132355233524147 is an evil number, because the sum of its binary digits is even.

The product of its digits is 9072000, while the sum is 50.

The spelling of 132355233524147 in words is "one hundred thirty-two trillion, three hundred fifty-five billion, two hundred thirty-three million, five hundred twenty-four thousand, one hundred forty-seven".