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134213242309 is a prime number
BaseRepresentation
bin111110011111110111…
…0111000110111000101
3110211102112220010112021
41330333232320313011
54144332042223214
6141353503323141
712460633510426
oct1747756706705
9424375803467
10134213242309
1151a12970473
12220178b04b1
13c86ba38b13
1466d2c8b64d
153757bab924
hex1f3fbb8dc5

134213242309 has 2 divisors, whose sum is σ = 134213242310. Its totient is φ = 134213242308.

The previous prime is 134213242291. The next prime is 134213242327. The reversal of 134213242309 is 903242312431.

It is a balanced prime because it is at equal distance from previous prime (134213242291) and next prime (134213242327).

It can be written as a sum of positive squares in only one way, i.e., 97314674209 + 36898568100 = 311953^2 + 192090^2 .

It is an emirp because it is prime and its reverse (903242312431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 134213242309 - 211 = 134213240261 is a prime.

It is a super-2 number, since 2×1342132423092 (a number of 23 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (134213242369) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67106621154 + 67106621155.

It is an arithmetic number, because the mean of its divisors is an integer number (67106621155).

Almost surely, 2134213242309 is an apocalyptic number.

It is an amenable number.

134213242309 is a deficient number, since it is larger than the sum of its proper divisors (1).

134213242309 is an equidigital number, since it uses as much as digits as its factorization.

134213242309 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 31104, while the sum is 34.

The spelling of 134213242309 in words is "one hundred thirty-four billion, two hundred thirteen million, two hundred forty-two thousand, three hundred nine".