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13014452000047 is a prime number
BaseRepresentation
bin1011110101100010100111…
…0010011101010100101111
31201002011120112111012100111
42331120221302131110233
53201212044203000142
643402431241040451
72512156152516331
oct275305162352457
951064515435314
1013014452000047
114168441960293
121562355786127
13735346b14434
1432dc8d55b451
15178808300e17
hexbd629c9d52f

13014452000047 has 2 divisors, whose sum is σ = 13014452000048. Its totient is φ = 13014452000046.

The previous prime is 13014451999937. The next prime is 13014452000063. The reversal of 13014452000047 is 74000025441031.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 13014452000047 - 211 = 13014451997999 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 13014451999982 and 13014452000018.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13014452000947) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6507226000023 + 6507226000024.

It is an arithmetic number, because the mean of its divisors is an integer number (6507226000024).

Almost surely, 213014452000047 is an apocalyptic number.

13014452000047 is a deficient number, since it is larger than the sum of its proper divisors (1).

13014452000047 is an equidigital number, since it uses as much as digits as its factorization.

13014452000047 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 13440, while the sum is 31.

Adding to 13014452000047 its reverse (74000025441031), we get a palindrome (87014477441078).

The spelling of 13014452000047 in words is "thirteen trillion, fourteen billion, four hundred fifty-two million, forty-seven", and thus it is an aban number.