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134071434353 is a prime number
BaseRepresentation
bin111110011011101000…
…1111011110001110001
3110211001200001112120202
41330313101323301301
54144034241344403
6141331440044545
712454262250233
oct1746721736161
9424050045522
10134071434353
115194a91400a
1221b98303755
13c84854789c
1466bc0d6253
15374a4e9688
hex1f3747bc71

134071434353 has 2 divisors, whose sum is σ = 134071434354. Its totient is φ = 134071434352.

The previous prime is 134071434307. The next prime is 134071434389. The reversal of 134071434353 is 353434170431.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 82962433024 + 51109001329 = 288032^2 + 226073^2 .

It is a cyclic number.

It is not a de Polignac number, because 134071434353 - 224 = 134054657137 is a prime.

It is a super-3 number, since 3×1340714343533 (a number of 34 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (134071434853) 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, 67035717176 + 67035717177.

It is an arithmetic number, because the mean of its divisors is an integer number (67035717177).

Almost surely, 2134071434353 is an apocalyptic number.

It is an amenable number.

134071434353 is a deficient number, since it is larger than the sum of its proper divisors (1).

134071434353 is an equidigital number, since it uses as much as digits as its factorization.

134071434353 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 181440, while the sum is 38.

The spelling of 134071434353 in words is "one hundred thirty-four billion, seventy-one million, four hundred thirty-four thousand, three hundred fifty-three".