Base | Representation |
---|---|
bin | 1011111011010010100001… |
… | …0011010011010011010010 |
3 | 1201102121111020112121101020 |
4 | 2332310220103103103102 |
5 | 3204321324043144442 |
6 | 43520043123420310 |
7 | 2522253412656540 |
oct | 276645023232322 |
9 | 51377436477336 |
10 | 13113211303122 |
11 | 41a6310a13427 |
12 | 1579517b24696 |
13 | 741755580888 |
14 | 33497b8d6d90 |
15 | 17b1886619ec |
hex | bed284d34d2 |
13113211303122 has 16 divisors (see below), whose sum is σ = 29973054407232. Its totient is φ = 3746631800880.
The previous prime is 13113211303091. The next prime is 13113211303139. The reversal of 13113211303122 is 22130311231131.
13113211303122 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a junction number, because it is equal to n+sod(n) for n = 13113211303092 and 13113211303101.
It is an unprimeable number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 156109658329 + ... + 156109658412.
It is an arithmetic number, because the mean of its divisors is an integer number (1873315900452).
Almost surely, 213113211303122 is an apocalyptic number.
13113211303122 is an abundant number, since it is smaller than the sum of its proper divisors (16859843104110).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
13113211303122 is a wasteful number, since it uses less digits than its factorization.
13113211303122 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 312219316753.
The product of its (nonzero) digits is 648, while the sum is 24.
Adding to 13113211303122 its reverse (22130311231131), we get a palindrome (35243522534253).
The spelling of 13113211303122 in words is "thirteen trillion, one hundred thirteen billion, two hundred eleven million, three hundred three thousand, one hundred twenty-two".
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