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1441212543571 is a prime number
BaseRepresentation
bin10100111110001110111…
…101100011001001010011
312002210000120122221001111
4110332032331203021103
5142103100402343241
63022030042314151
7206060406331213
oct24761675431123
95083016587044
101441212543571
1150623a9aa259
121b33971b8957
13a5ba0c53128
144da7dc50243
152775130dc81
hex14f8ef63253

1441212543571 has 2 divisors, whose sum is σ = 1441212543572. Its totient is φ = 1441212543570.

The previous prime is 1441212543517. The next prime is 1441212543619. The reversal of 1441212543571 is 1753452121441.

Together with previous prime (1441212543517) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It is an emirp because it is prime and its reverse (1753452121441) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1441212543571 - 213 = 1441212535379 is a prime.

It is not a weakly prime, because it can be changed into another prime (1441212543271) 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, 720606271785 + 720606271786.

It is an arithmetic number, because the mean of its divisors is an integer number (720606271786).

Almost surely, 21441212543571 is an apocalyptic number.

1441212543571 is a deficient number, since it is larger than the sum of its proper divisors (1).

1441212543571 is an equidigital number, since it uses as much as digits as its factorization.

1441212543571 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 134400, while the sum is 40.

The spelling of 1441212543571 in words is "one trillion, four hundred forty-one billion, two hundred twelve million, five hundred forty-three thousand, five hundred seventy-one".