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13163606130539 is a prime number
BaseRepresentation
bin1011111110001110010000…
…0100010011111101101011
31201121102120100022002101202
42333203210010103331223
53211133031132134124
643555135511434415
72526016300641614
oct277434404237553
951542510262352
1013163606130539
114215721539127
121587241085a0b
13746427070697
1433719c842c0b
1517c6379e75ae
hexbf8e4113f6b

13163606130539 has 2 divisors, whose sum is σ = 13163606130540. Its totient is φ = 13163606130538.

The previous prime is 13163606130463. The next prime is 13163606130541. The reversal of 13163606130539 is 93503160636131.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-13163606130539 is a prime.

Together with 13163606130541, 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 = 13163606130493 and 13163606130502.

It is not a weakly prime, because it can be changed into another prime (13163606130599) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6581803065269 + 6581803065270.

It is an arithmetic number, because the mean of its divisors is an integer number (6581803065270).

Almost surely, 213163606130539 is an apocalyptic number.

13163606130539 is a deficient number, since it is larger than the sum of its proper divisors (1).

13163606130539 is an equidigital number, since it uses as much as digits as its factorization.

13163606130539 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 787320, while the sum is 47.

The spelling of 13163606130539 in words is "thirteen trillion, one hundred sixty-three billion, six hundred six million, one hundred thirty thousand, five hundred thirty-nine".