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12203221300147 is a prime number
BaseRepresentation
bin1011000110010100100010…
…1010111011001110110011
31121012121122110210210100201
42301211020222323032303
53044414144123101042
641542025451500031
72366440125402103
oct261451052731663
947177573723321
1012203221300147
1139853a2317529
121451096157617
136a69b31c7707
143028d3831203
1516267909e1b7
hexb1948abb3b3

12203221300147 has 2 divisors, whose sum is σ = 12203221300148. Its totient is φ = 12203221300146.

The previous prime is 12203221300133. The next prime is 12203221300253. The reversal of 12203221300147 is 74100312230221.

It is a weak prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-12203221300147 is a prime.

It is a super-2 number, since 2×122032213001472 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (12203221400147) 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, 6101610650073 + 6101610650074.

It is an arithmetic number, because the mean of its divisors is an integer number (6101610650074).

Almost surely, 212203221300147 is an apocalyptic number.

12203221300147 is a deficient number, since it is larger than the sum of its proper divisors (1).

12203221300147 is an equidigital number, since it uses as much as digits as its factorization.

12203221300147 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 4032, while the sum is 28.

Adding to 12203221300147 its reverse (74100312230221), we get a palindrome (86303533530368).

The spelling of 12203221300147 in words is "twelve trillion, two hundred three billion, two hundred twenty-one million, three hundred thousand, one hundred forty-seven".