Base | Representation |
---|---|
bin | 110100100011011000100111… |
… | …011111000111001100100001 |
3 | 1010022100211000100220101211122 |
4 | 310203120213133013030201 |
5 | 220243313301312044210 |
6 | 2135323400543422025 |
7 | 66453410446042442 |
oct | 6443304737071441 |
9 | 1108324010811748 |
10 | 231130032534305 |
11 | 677107728aa983 |
12 | 21b0a6702b4915 |
13 | 9bc761251cb48 |
14 | 4110a78d512c9 |
15 | 1bac35121cb55 |
hex | d236277c7321 |
231130032534305 has 16 divisors (see below), whose sum is σ = 277392525004800. Its totient is φ = 184879703616960.
The previous prime is 231130032534287. The next prime is 231130032534311. The reversal of 231130032534305 is 503435230031132.
231130032534305 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 231130032534305 - 224 = 231130015757089 is a prime.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 1450730600 + ... + 1450889910.
It is an arithmetic number, because the mean of its divisors is an integer number (17337032812800).
Almost surely, 2231130032534305 is an apocalyptic number.
It is an amenable number.
231130032534305 is a deficient number, since it is larger than the sum of its proper divisors (46262492470495).
231130032534305 is a wasteful number, since it uses less digits than its factorization.
231130032534305 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 195664.
The product of its (nonzero) digits is 97200, while the sum is 35.
Adding to 231130032534305 its reverse (503435230031132), we get a palindrome (734565262565437).
The spelling of 231130032534305 in words is "two hundred thirty-one trillion, one hundred thirty billion, thirty-two million, five hundred thirty-four thousand, three hundred five".
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