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1338144260287 is a prime number
BaseRepresentation
bin10011011110001111100…
…111010000000010111111
311201220222120111122102011
4103132033213100002333
5133411004412312122
62502422431312051
7165451314353332
oct23361747200277
94656876448364
101338144260287
1147655a60554a
1219741197b627
1399256871b63
1448aa34d0619
1524c1ca43377
hex1378f9d00bf

1338144260287 has 2 divisors, whose sum is σ = 1338144260288. Its totient is φ = 1338144260286.

The previous prime is 1338144260281. The next prime is 1338144260291. The reversal of 1338144260287 is 7820624418331.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 1338144260287 - 23 = 1338144260279 is a prime.

It is a super-2 number, since 2×13381442602872 (a number of 25 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 1338144260287.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1338144260281) 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, 669072130143 + 669072130144.

It is an arithmetic number, because the mean of its divisors is an integer number (669072130144).

Almost surely, 21338144260287 is an apocalyptic number.

1338144260287 is a deficient number, since it is larger than the sum of its proper divisors (1).

1338144260287 is an equidigital number, since it uses as much as digits as its factorization.

1338144260287 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1548288, while the sum is 49.

The spelling of 1338144260287 in words is "one trillion, three hundred thirty-eight billion, one hundred forty-four million, two hundred sixty thousand, two hundred eighty-seven".