Base | Representation |
---|---|
bin | 11001101110000111100111… |
… | …010000010111100011000000 |
3 | 112211112012121111200212120002 |
4 | 121232013213100113203000 |
5 | 104311331121434034100 |
6 | 1040330450523135132 |
7 | 32553452536614512 |
oct | 3156074720274300 |
9 | 484465544625502 |
10 | 113120433502400 |
11 | 33053119739835 |
12 | 1082b5b83744a8 |
13 | 4b17298a5a1cc |
14 | 1dd10b5985bb2 |
15 | d127c6edc2d5 |
hex | 66e1e74178c0 |
113120433502400 has 84 divisors (see below), whose sum is σ = 294720317739288. Its totient is φ = 42586516131840.
The previous prime is 113120433502397. The next prime is 113120433502421. The reversal of 113120433502400 is 4205334021311.
It is an unprimeable number.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 2079392534 + ... + 2079446933.
It is an arithmetic number, because the mean of its divisors is an integer number (3508575211182).
Almost surely, 2113120433502400 is an apocalyptic number.
113120433502400 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
113120433502400 is an abundant number, since it is smaller than the sum of its proper divisors (181599884236888).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
113120433502400 is a wasteful number, since it uses less digits than its factorization.
113120433502400 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 4158839506 (or 4158839491 counting only the distinct ones).
The product of its (nonzero) digits is 8640, while the sum is 29.
Adding to 113120433502400 its reverse (4205334021311), we get a palindrome (117325767523711).
The spelling of 113120433502400 in words is "one hundred thirteen trillion, one hundred twenty billion, four hundred thirty-three million, five hundred two thousand, four hundred".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.070 sec. • engine limits •