101144121300 has 432 divisors, whose sum is σ = 332685133872. Its totient is φ = 25679462400.

The previous prime is 101144121277. The next prime is 101144121301. The reversal of 101144121300 is 3121441101.

101144121300 is a `hidden beast` number, since 1 + 0 + 11 + 441 + 213 + 0 + 0 = 666.

It can be written as a sum of positive squares in 24 ways, for example, as 54607277124 + 46536844176 = 233682^2 + 215724^2 .

It is a super-2 number, since 2×101144121300² (a number of 23 digits) contains 22 as substring.

It is a Harshad number since it is a multiple of its sum of digits (18).

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (101144121301) by changing a digit.

It is a polite number, since it can be written in 143 ways as a sum of consecutive naturals, for example, 558807210 + ... + 558807390.

Almost surely, 2^101144121300 is an apocalyptic number.

101144121300 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 101144121300, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (166342566936).

101144121300 is an abundant number, since it is smaller than the sum of its proper divisors (231541012572).

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

101144121300 is a wasteful number, since it uses less digits than its factorization.

101144121300 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 508 (or 498 counting only the distinct ones).

The product of its (nonzero) digits is 96, while the sum is 18.

Adding to 101144121300 its reverse (3121441101), we get a palindrome (104265562401).

The spelling of 101144121300 in words is "one hundred one billion, one hundred forty-four million, one hundred twenty-one thousand, three hundred".