1151214433 has 4 divisors (see below), whose sum is σ = 1151628820. Its totient is φ = 1150800048.

The previous prime is 1151214413. The next prime is 1151214437. The reversal of 1151214433 is 3344121511.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in 2 ways, for example, as 307195729 + 844018704 = 17527^2 + 29052^2 .

It is a cyclic number.

It is not a de Polignac number, because 1151214433 - 2⁵ = 1151214401 is a prime.

It is a super-2 number, since 2×1151214433² = 2650589341495022978, which contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 1151214398 and 1151214407.

It is not an unprimeable number, because it can be changed into a prime (1151214437) 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 in 3 ways as a sum of consecutive naturals, for example, 202998 + ... + 208591.

It is an arithmetic number, because the mean of its divisors is an integer number (287907205).

Almost surely, 2^1151214433 is an apocalyptic number.

It is an amenable number.

1151214433 is a deficient number, since it is larger than the sum of its proper divisors (414387).

1151214433 is an equidigital number, since it uses as much as digits as its factorization.

1151214433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 414386.

The product of its digits is 1440, while the sum is 25.

The square root of 1151214433 is about 33929.5510285651. The cubic root of 1151214433 is about 1048.0582201719.

Adding to 1151214433 its reverse (3344121511), we get a palindrome (4495335944).

The spelling of 1151214433 in words is "one billion, one hundred fifty-one million, two hundred fourteen thousand, four hundred thirty-three".