1131243433 has 4 divisors (see below), whose sum is σ = 1132519680. Its totient is φ = 1129967188.

The previous prime is 1131243401. The next prime is 1131243457. The reversal of 1131243433 is 3343421311.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also an emirpimes, since its reverse is a distinct semiprime: 3343421311 = 71 ⋅47090441.

It is a cyclic number.

It is not a de Polignac number, because 1131243433 - 2⁵ = 1131243401 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 1131243398 and 1131243407.

It is not an unprimeable number, because it can be changed into a prime (1131243473) 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 in 3 ways as a sum of consecutive naturals, for example, 636793 + ... + 638566.

It is an arithmetic number, because the mean of its divisors is an integer number (283129920).

Almost surely, 2^1131243433 is an apocalyptic number.

It is an amenable number.

1131243433 is a deficient number, since it is larger than the sum of its proper divisors (1276247).

1131243433 is an equidigital number, since it uses as much as digits as its factorization.

1131243433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1276246.

The product of its digits is 2592, while the sum is 25.

The square root of 1131243433 is about 33633.9624932894. The cubic root of 1131243433 is about 1041.9623431265.

Adding to 1131243433 its reverse (3343421311), we get a palindrome (4474664744).

It can be divided in two parts, 11312 and 43433, that added together give a palindrome (54745).

The spelling of 1131243433 in words is "one billion, one hundred thirty-one million, two hundred forty-three thousand, four hundred thirty-three".