Search a number
-
+
133020525134447 is a prime number
BaseRepresentation
bin11110001111101101000001…
…001001100000001001101111
3122102222122020002221102111122
4132033231001021200021233
5114413402013413300242
61150524441410211155
740006262421551636
oct3617550111401157
9572878202842448
10133020525134447
1139425762667235
1212b04328883abb
13592ba21505252
1424bc328dd651d
15105a280bd8ad2
hex78fb4126026f

133020525134447 has 2 divisors, whose sum is σ = 133020525134448. Its totient is φ = 133020525134446.

The previous prime is 133020525134411. The next prime is 133020525134449. The reversal of 133020525134447 is 744431525020331.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 133020525134447 - 232 = 133016230167151 is a prime.

Together with 133020525134449, 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 = 133020525134398 and 133020525134407.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (133020525134449) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66510262567223 + 66510262567224.

It is an arithmetic number, because the mean of its divisors is an integer number (66510262567224).

Almost surely, 2133020525134447 is an apocalyptic number.

133020525134447 is a deficient number, since it is larger than the sum of its proper divisors (1).

133020525134447 is an equidigital number, since it uses as much as digits as its factorization.

133020525134447 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1209600, while the sum is 44.

The spelling of 133020525134447 in words is "one hundred thirty-three trillion, twenty billion, five hundred twenty-five million, one hundred thirty-four thousand, four hundred forty-seven".