1340203433 has 2 divisors, whose sum is σ = 1340203434. Its totient is φ = 1340203432.

The previous prime is 1340203427. The next prime is 1340203453. The reversal of 1340203433 is 3343020431.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1173199504 + 167003929 = 34252^2 + 12923^2 .

It is an emirp because it is prime and its reverse (3343020431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1340203433 - 2¹⁸ = 1339941289 is a prime.

It is a super-2 number, since 2×1340203433² = 3592290483649970978, which contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1340203399 and 1340203408.

It is not a weakly prime, because it can be changed into another prime (1340203453) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 670101716 + 670101717.

It is an arithmetic number, because the mean of its divisors is an integer number (670101717).

Almost surely, 2^1340203433 is an apocalyptic number.

It is an amenable number.

1340203433 is a deficient number, since it is larger than the sum of its proper divisors (1).

1340203433 is an equidigital number, since it uses as much as digits as its factorization.

1340203433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 2592, while the sum is 23.

The square root of 1340203433 is about 36608.7890130225. The cubic root of 1340203433 is about 1102.5295595587.

Adding to 1340203433 its reverse (3343020431), we get a palindrome (4683223864).

The spelling of 1340203433 in words is "one billion, three hundred forty million, two hundred three thousand, four hundred thirty-three".