Base | Representation |
---|---|
bin | 11110101110010100000… |
… | …000100100100110001011 |
3 | 21110211122222020102212201 |
4 | 132232110000210212023 |
5 | 234042431400321303 |
6 | 4253531242021031 |
7 | 305352165625606 |
oct | 36562400444613 |
9 | 7424588212781 |
10 | 2111312120203 |
11 | 744447717260 |
12 | 2a122a1a7777 |
13 | 124132850b60 |
14 | 7428c02b33d |
15 | 39dc03ab81d |
hex | 1eb9402498b |
2111312120203 has 32 divisors (see below), whose sum is σ = 2611161181440. Its totient is φ = 1678359346560.
The previous prime is 2111312120189. The next prime is 2111312120207. The reversal of 2111312120203 is 3020212131112.
It is not a de Polignac number, because 2111312120203 - 25 = 2111312120171 is a prime.
It is a super-2 number, since 2×21113121202032 (a number of 25 digits) contains 22 as substring.
It is a Harshad number since it is a multiple of its sum of digits (19).
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (2111312120207) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 63987876 + ... + 64020862.
It is an arithmetic number, because the mean of its divisors is an integer number (81598786920).
Almost surely, 22111312120203 is an apocalyptic number.
2111312120203 is a deficient number, since it is larger than the sum of its proper divisors (499849061237).
2111312120203 is a wasteful number, since it uses less digits than its factorization.
2111312120203 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 56587.
The product of its (nonzero) digits is 144, while the sum is 19.
Adding to 2111312120203 its reverse (3020212131112), we get a palindrome (5131524251315).
The spelling of 2111312120203 in words is "two trillion, one hundred eleven billion, three hundred twelve million, one hundred twenty thousand, two hundred three".
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