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13140310121171 is a prime number
BaseRepresentation
bin1011111100110111011110…
…0001000101101011010011
31201112012102210202012021222
42333031313201011223103
53210242323342334141
643540324121235255
72524233064063142
oct277156741055323
951465383665258
1013140310121171
114206857550522
12158281b332b2b
1374418387cc5c
14335dcc8dd759
1517bc2271614b
hexbf377845ad3

13140310121171 has 2 divisors, whose sum is σ = 13140310121172. Its totient is φ = 13140310121170.

The previous prime is 13140310121111. The next prime is 13140310121173. The reversal of 13140310121171 is 17112101304131.

It is a happy number.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-13140310121171 is a prime.

It is a super-2 number, since 2×131403101211712 (a number of 27 digits) contains 22 as substring.

Together with 13140310121173, it forms a pair of twin primes.

It is a Chen prime.

It is a self number, because there is not a number n which added to its sum of digits gives 13140310121171.

It is not a weakly prime, because it can be changed into another prime (13140310121173) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6570155060585 + 6570155060586.

It is an arithmetic number, because the mean of its divisors is an integer number (6570155060586).

Almost surely, 213140310121171 is an apocalyptic number.

13140310121171 is a deficient number, since it is larger than the sum of its proper divisors (1).

13140310121171 is an equidigital number, since it uses as much as digits as its factorization.

13140310121171 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 504, while the sum is 26.

The spelling of 13140310121171 in words is "thirteen trillion, one hundred forty billion, three hundred ten million, one hundred twenty-one thousand, one hundred seventy-one".