Base | Representation |
---|---|
bin | 11011100010011011010010… |
… | …101001111001011111000010 |
3 | 120212211021212202001112000010 |
4 | 123202123102221321133002 |
5 | 111333303432113330310 |
6 | 1105330315413205350 |
7 | 34340061406163256 |
oct | 3342332251713702 |
9 | 525737782045003 |
10 | 121113022011330 |
11 | 35654820574038 |
12 | 11700622030856 |
13 | 5276ba801269b |
14 | 21c9c89b92266 |
15 | e00663455320 |
hex | 6e26d2a797c2 |
121113022011330 has 128 divisors (see below), whose sum is σ = 304706855462400. Its totient is φ = 30750440269824.
The previous prime is 121113022011301. The next prime is 121113022011349. The reversal of 121113022011330 is 33110220311121.
121113022011330 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is a Curzon number.
It is a junction number, because it is equal to n+sod(n) for n = 121113022011297 and 121113022011306.
It is an unprimeable number.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 209264712 + ... + 209842668.
It is an arithmetic number, because the mean of its divisors is an integer number (2380522308300).
Almost surely, 2121113022011330 is an apocalyptic number.
121113022011330 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is a practical number, because each smaller number is the sum of distinct divisors of 121113022011330, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (152353427731200).
121113022011330 is an abundant number, since it is smaller than the sum of its proper divisors (183593833451070).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
121113022011330 is a wasteful number, since it uses less digits than its factorization.
121113022011330 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 579388.
The product of its (nonzero) digits is 216, while the sum is 21.
Adding to 121113022011330 its reverse (33110220311121), we get a palindrome (154223242322451).
The spelling of 121113022011330 in words is "one hundred twenty-one trillion, one hundred thirteen billion, twenty-two million, eleven thousand, three hundred thirty".
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