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134100433213 is a prime number
BaseRepresentation
bin111110011100100000…
…0100011100100111101
3110211010200121211112121
41330321000203210331
54144114202330323
6141334353402541
712455063606065
oct1747100434475
9424120554477
10134100433213
115196522030a
1221ba5b69451
13c85155bc86
1466c1cc43a5
15374cd26a5d
hex1f3902393d

134100433213 has 2 divisors, whose sum is σ = 134100433214. Its totient is φ = 134100433212.

The previous prime is 134100433183. The next prime is 134100433217. The reversal of 134100433213 is 312334001431.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 105055070884 + 29045362329 = 324122^2 + 170427^2 .

It is an emirp because it is prime and its reverse (312334001431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 134100433213 - 233 = 125510498621 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (134100433217) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67050216606 + 67050216607.

It is an arithmetic number, because the mean of its divisors is an integer number (67050216607).

Almost surely, 2134100433213 is an apocalyptic number.

It is an amenable number.

134100433213 is a deficient number, since it is larger than the sum of its proper divisors (1).

134100433213 is an equidigital number, since it uses as much as digits as its factorization.

134100433213 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 2592, while the sum is 25.

Adding to 134100433213 its reverse (312334001431), we get a palindrome (446434434644).

The spelling of 134100433213 in words is "one hundred thirty-four billion, one hundred million, four hundred thirty-three thousand, two hundred thirteen".