130010011000032 has 192 divisors, whose sum is σ = 347299709460480. Its totient is φ = 42578692079616.

The previous prime is 130010011000019. The next prime is 130010011000049. The reversal of 130010011000032 is 230000110010031.

130010011000032 is digitally balanced in base 3, 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 (12).

It is a junction number, because it is equal to n+sod(n) for n = 130010010999981 and 130010011000017.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 4092338644 + ... + 4092370412.

It is an arithmetic number, because the mean of its divisors is an integer number (1808852653440).

Almost surely, 2^{130010011000032} is an apocalyptic number.

130010011000032 is a gapful number since it is divisible by the number (12) 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 130010011000032, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (173649854730240).

130010011000032 is an abundant number, since it is smaller than the sum of its proper divisors (217289698460448).

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

130010011000032 is a wasteful number, since it uses less digits than its factorization.

130010011000032 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 35153 (or 35145 counting only the distinct ones).

The product of its (nonzero) digits is 18, while the sum is 12.

Adding to 130010011000032 its reverse (230000110010031), we get a palindrome (360010121010063).

Subtracting 130010011000032 from its reverse (230000110010031), we obtain a palindrome (99990099009999).

The spelling of 130010011000032 in words is "one hundred thirty trillion, ten billion, eleven million, thirty-two", and thus it is an aban number.