433400113120 has 96 divisors (see below), whose sum is σ = 1103018757120. Its totient is φ = 159974031360.

The previous prime is 433400113063. The next prime is 433400113169. The reversal of 433400113120 is 21311004334.

It is a super-2 number, since 2×433400113120² (a number of 24 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 433400113091 and 433400113100.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 6893025 + ... + 6955615.

It is an arithmetic number, because the mean of its divisors is an integer number (11489778720).

Almost surely, 2^433400113120 is an apocalyptic number.

433400113120 is a gapful number since it is divisible by the number (40) formed by its first and last digit.

It is an amenable number.

It is a practical number, because each smaller number is the sum of distinct divisors of 433400113120, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (551509378560).

433400113120 is an abundant number, since it is smaller than the sum of its proper divisors (669618644000).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

433400113120 is a wasteful number, since it uses less digits than its factorization.

433400113120 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 65948 (or 65940 counting only the distinct ones).

The product of its (nonzero) digits is 864, while the sum is 22.

Adding to 433400113120 its reverse (21311004334), we get a palindrome (454711117454).

The spelling of 433400113120 in words is "four hundred thirty-three billion, four hundred million, one hundred thirteen thousand, one hundred twenty".