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24312143131489 is a prime number
BaseRepresentation
bin1011000011100100111001…
…01001111011001101100001
310012002012220001112020011201
411201302130221323031201
511141312232120201424
6123412502214441201
75056331366616163
oct541623451731541
9105065801466151
1024312143131489
1178237a3a88364
122887a33663201
131074818908375
146009dd338933
152c26345e4d44
hex161c9ca7b361

24312143131489 has 2 divisors, whose sum is σ = 24312143131490. Its totient is φ = 24312143131488.

The previous prime is 24312143131439. The next prime is 24312143131499. The reversal of 24312143131489 is 98413134121342.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 24238045936225 + 74097195264 = 4923215^2 + 272208^2 .

It is a cyclic number.

It is not a de Polignac number, because 24312143131489 - 225 = 24312109577057 is a prime.

It is a super-3 number, since 3×243121431314893 (a number of 41 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (24312143131439) 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, 12156071565744 + 12156071565745.

It is an arithmetic number, because the mean of its divisors is an integer number (12156071565745).

Almost surely, 224312143131489 is an apocalyptic number.

It is an amenable number.

24312143131489 is a deficient number, since it is larger than the sum of its proper divisors (1).

24312143131489 is an equidigital number, since it uses as much as digits as its factorization.

24312143131489 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 497664, while the sum is 46.

The spelling of 24312143131489 in words is "twenty-four trillion, three hundred twelve billion, one hundred forty-three million, one hundred thirty-one thousand, four hundred eighty-nine".