11313402433 has 2 divisors, whose sum is σ = 11313402434. Its totient is φ = 11313402432.

The previous prime is 11313402413. The next prime is 11313402461. The reversal of 11313402433 is 33420431311.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 7883308944 + 3430093489 = 88788^2 + 58567^2 .

It is a cyclic number.

It is not a de Polignac number, because 11313402433 - 2⁹ = 11313401921 is a prime.

It is a super-2 number, since 2×11313402433² (a number of 21 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 11313402398 and 11313402407.

It is not a weakly prime, because it can be changed into another prime (11313402413) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5656701216 + 5656701217.

It is an arithmetic number, because the mean of its divisors is an integer number (5656701217).

Almost surely, 2^11313402433 is an apocalyptic number.

It is an amenable number.

11313402433 is a deficient number, since it is larger than the sum of its proper divisors (1).

11313402433 is an equidigital number, since it uses as much as digits as its factorization.

11313402433 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 25.

Adding to 11313402433 its reverse (33420431311), we get a palindrome (44733833744).

The spelling of 11313402433 in words is "eleven billion, three hundred thirteen million, four hundred two thousand, four hundred thirty-three".