Base | Representation |
---|---|
bin | 11001001010110… |
… | …01101110100100 |
3 | 112201021120200210 |
4 | 30211121232210 |
5 | 413022200200 |
6 | 32541135420 |
7 | 5142405102 |
oct | 1445315644 |
9 | 481246623 |
10 | 211131300 |
11 | a91a6086 |
12 | 5a85a570 |
13 | 34983a83 |
14 | 2007cc72 |
15 | 13807650 |
hex | c959ba4 |
211131300 has 72 divisors (see below), whose sum is σ = 612412192. Its totient is φ = 56160000.
The previous prime is 211131289. The next prime is 211131307. The reversal of 211131300 is 3131112.
211131300 is digitally balanced in base 2 and base 3, because in such bases 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 not an unprimeable number, because it can be changed into a prime (211131307) 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, 179715 + ... + 180885.
Almost surely, 2211131300 is an apocalyptic number.
211131300 is a gapful number since it is divisible by the number (20) 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 211131300, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (306206096).
211131300 is an abundant number, since it is smaller than the sum of its proper divisors (401280892).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
211131300 is a wasteful number, since it uses less digits than its factorization.
211131300 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1789 (or 1782 counting only the distinct ones).
The product of its (nonzero) digits is 18, while the sum is 12.
The square root of 211131300 is about 14530.3578758405. The cubic root of 211131300 is about 595.4576428775.
Adding to 211131300 its reverse (3131112), we get a palindrome (214262412).
It can be divided in two parts, 2111 and 31300, that added together give a triangular number (33411 = T258).
The spelling of 211131300 in words is "two hundred eleven million, one hundred thirty-one thousand, three hundred".
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