13011311433120 has 1152 divisors, whose sum is σ = 56828488550400. Its totient is φ = 2428182282240.

The previous prime is 13011311433103. The next prime is 13011311433173. The reversal of 13011311433120 is 2133411311031.

13011311433120 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a Harshad number since it is a multiple of its sum of digits (24).

It is an unprimeable number.

It is a polite number, since it can be written in 191 ways as a sum of consecutive naturals, for example, 231082731 + ... + 231139029.

It is an arithmetic number, because the mean of its divisors is an integer number (49330285200).

Almost surely, 2^{13011311433120} is an apocalyptic number.

13011311433120 is a gapful number since it is divisible by the number (10) 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 13011311433120, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (28414244275200).

13011311433120 is an abundant number, since it is smaller than the sum of its proper divisors (43817177117280).

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

13011311433120 is a wasteful number, since it uses less digits than its factorization.

13011311433120 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 56398 (or 56377 counting only the distinct ones).

The product of its (nonzero) digits is 648, while the sum is 24.

Adding to 13011311433120 its reverse (2133411311031), we get a palindrome (15144722744151).

It can be divided in two parts, 13011311 and 433120, that added together give a palindrome (13444431).

The spelling of 13011311433120 in words is "thirteen trillion, eleven billion, three hundred eleven million, four hundred thirty-three thousand, one hundred twenty".