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1335531242849 is a prime number
BaseRepresentation
bin10011011011110011110…
…111011000010101100001
311201200020110122121100112
4103123303313120111201
5133340131444232344
62501311245301105
7165326465154566
oct23336367302541
94650213577315
101335531242849
11475439629901
12196a02864795
1398c2b3c1aab
14488d646086d
1524b18440c9e
hex136f3dd8561

1335531242849 has 2 divisors, whose sum is σ = 1335531242850. Its totient is φ = 1335531242848.

The previous prime is 1335531242767. The next prime is 1335531242863. The reversal of 1335531242849 is 9482421355331.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 745295523025 + 590235719824 = 863305^2 + 768268^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1335531242849 is a prime.

It is a super-2 number, since 2×13355312428492 (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1335531242797 and 1335531242806.

It is not a weakly prime, because it can be changed into another prime (1335531242879) 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, 667765621424 + 667765621425.

It is an arithmetic number, because the mean of its divisors is an integer number (667765621425).

Almost surely, 21335531242849 is an apocalyptic number.

It is an amenable number.

1335531242849 is a deficient number, since it is larger than the sum of its proper divisors (1).

1335531242849 is an equidigital number, since it uses as much as digits as its factorization.

1335531242849 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 3110400, while the sum is 50.

The spelling of 1335531242849 in words is "one trillion, three hundred thirty-five billion, five hundred thirty-one million, two hundred forty-two thousand, eight hundred forty-nine".