11312115433 has 4 divisors (see below), whose sum is σ = 11313200340. Its totient is φ = 11311030528.

The previous prime is 11312115431. The next prime is 11312115449. The reversal of 11312115433 is 33451121311.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 33451121311 = 139 ⋅240655549.

It can be written as a sum of positive squares in 2 ways, for example, as 8023322329 + 3288793104 = 89573^2 + 57348^2 .

It is a cyclic number.

It is not a de Polignac number, because 11312115433 - 2¹ = 11312115431 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 11312115398 and 11312115407.

It is not an unprimeable number, because it can be changed into a prime (11312115431) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 526660 + ... + 547717.

It is an arithmetic number, because the mean of its divisors is an integer number (2828300085).

Almost surely, 2^11312115433 is an apocalyptic number.

It is an amenable number.

11312115433 is a deficient number, since it is larger than the sum of its proper divisors (1084907).

11312115433 is a wasteful number, since it uses less digits than its factorization.

11312115433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1084906.

The product of its digits is 1080, while the sum is 25.

Adding to 11312115433 its reverse (33451121311), we get a palindrome (44763236744).

The spelling of 11312115433 in words is "eleven billion, three hundred twelve million, one hundred fifteen thousand, four hundred thirty-three".