Base | Representation |
---|---|
bin | 110100101101101000… |
… | …0001010001010101100 |
3 | 101211012002222112011212 |
4 | 1221123100022022230 |
5 | 3323313212043200 |
6 | 124000423133552 |
7 | 11115130035632 |
oct | 1513320121254 |
9 | 354162875155 |
10 | 113200112300 |
11 | 4400a56a920 |
12 | 19b325a42b8 |
13 | a8a04a7892 |
14 | 569c1db352 |
15 | 2e28019035 |
hex | 1a5b40a2ac |
113200112300 has 72 divisors (see below), whose sum is σ = 268164503040. Its totient is φ = 41134651200.
The previous prime is 113200112297. The next prime is 113200112303. The reversal of 113200112300 is 3211002311.
It is an interprime number because it is at equal distance from previous prime (113200112297) and next prime (113200112303).
It is a super-2 number, since 2×1132001123002 (a number of 23 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (113200112303) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 1556141 + ... + 1627259.
Almost surely, 2113200112300 is an apocalyptic number.
113200112300 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 113200112300, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (134082251520).
113200112300 is an abundant number, since it is smaller than the sum of its proper divisors (154964390740).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
113200112300 is a wasteful number, since it uses less digits than its factorization.
113200112300 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 72591 (or 72584 counting only the distinct ones).
The product of its (nonzero) digits is 36, while the sum is 14.
Adding to 113200112300 its reverse (3211002311), we get a palindrome (116411114611).
It can be divided in two parts, 11320011 and 2300, that added together give a palindrome (11322311).
The spelling of 113200112300 in words is "one hundred thirteen billion, two hundred million, one hundred twelve thousand, three hundred".
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