Empty
  • Empty
Number with only two factors.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Empty
  • Empty
If you're given \(n= 2^a 3^b\) then can you know what the next largest number is: \(m=2^x3^y\)? Basically, given a,b can we find x,y?
Empty
  • Empty
Here is an example: \(2 = 2^1 3^0\) so the next number is \(3=2^03^1\). Here's the start of the trend: \(2 < 3 < 4 < 6 < 8 < 9 < \cdots\)
UsukiDoll
  • UsukiDoll
that's when n = 2 a = 1 and b = 0 for n =2^a3^b I guess the same trend will be for m = 2^x3^y too?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Empty
  • Empty
Well it's simple to see the trend within numbers that have the same number of prime factors, for instance: \(2*2<2*3<3*3\) Which is the same as saying \(4 < 6 <9 \) However there is overlap sometimes, when does this occur? Afterall, we have: \(2*3 < 2*2*2 < 3*3\) This is a simple example \(6<8<9\) So is this kind of behavior predictable?
UsukiDoll
  • UsukiDoll
can n and m be a number besides 2 and 3?
Empty
  • Empty
Yeah, they are just consecutive numbers that have only factors of 2 and 3.
UsukiDoll
  • UsukiDoll
\[4= 2^1 3^0\] \[5= 2^0 3^1\] like this?
UsukiDoll
  • UsukiDoll
NUGH!!!!!!!!!!!! I don't know how to start this afterwards
Empty
  • Empty
Oh ok well first off, wait I'm confused, these are just exponents: \(4 =2^2 3^0 \ne 2^1 3^0 = 2\) \(5 \ne 2^03^1 = 3\) So n and m depends on the exponents: \(2^13^0 < 2^0 3^1 < 2^23^0 < 2^1 3^1\) So this is the same as \(2 <3 <4 <6 \)
UsukiDoll
  • UsukiDoll
wow. 5 got skipped.
UsukiDoll
  • UsukiDoll
maybe \[2^03^2 =1(9) =9 \] \[2^23^1 =4(3) =12 \] \[2^13^2 =2(9) =18 \] \[2^23^2 =4(9) =36 \]
Empty
  • Empty
There is no 5, only factors of 2 and 3, and you can't make 5 by multiplying 2 and 3 together so it's not included. :P
UsukiDoll
  • UsukiDoll
yeah I know... 2 x 3 or 2^13^1 = 2(3) = 6
UsukiDoll
  • UsukiDoll
so with the exponent numbers we can only produce certain results. like \[2^13^1 = (2)(3) = 6\] \[2^13^2 = 2(9) = 18 \] but I'm getting big numbers XD
UsukiDoll
  • UsukiDoll
like let 2 be fixed with that exponent 1 only and let 3 have the b = 1,2,3,4,5,............... \[2^13^3 = 2(27) = 54\] \[2^13^4 = 2(81) = 162\] \[2^13^5 = 2(243) = 486\]
UsukiDoll
  • UsukiDoll
oh I see what I did wrong earlier XD I just assigned m and n without thinking about a and b. That's why \[4 \neq 2 , 5 \neq 3 \]
Empty
  • Empty
Yeah! Now you understand where I'm stuck now haha. I just sorta came up with this problem on my own earlier and I forget why but it's not the first time I've seen this before so it must be important to me.
UsukiDoll
  • UsukiDoll
if all of the exponents for 2 were 0 \[2^03^3 = 1(27) =27......................2^03^4 =(1)(81) = 81..........2^03^5=(1)(243) = 243\]
UsukiDoll
  • UsukiDoll
it's like if the exponent number is bigger for 3, then we have more threes if we write it out... it dominates or something like that. like \[2^13^3 = 2(3)(3)(3) =54\]
ganeshie8
  • ganeshie8
\[\large n= 2^a 3^b = 2^{a+b\log_2 3}\]
UsukiDoll
  • UsukiDoll
-_- now the problem *at least for me* looks a bit complicated... in that log form
UsukiDoll
  • UsukiDoll
unless a = 1 and b = 1
ganeshie8
  • ganeshie8
Not really, notice that we no longer have to worry about two different bases
UsukiDoll
  • UsukiDoll
\[\large n= 2^a 3^b = 2^{a+b\log_2 3}\] if a = 1 and b = 1 \[\large n= 2^1 3^1 = 2^{1+1\log_2 3}\]
UsukiDoll
  • UsukiDoll
nah I prefer the different bases.
ganeshie8
  • ganeshie8
The problem translates to finding the smallest positive integers \(x,y\) given \(a, b\) that satisfy : \( a + b\log_2 3\le x+y\log_2 3 \)
ganeshie8
  • ganeshie8
at least the logs give us some alternative but concrete thing to mess with..
ganeshie8
  • ganeshie8
multiplying \(\log 2\) through out that becomes \( a\log 2 + b\log 3\le x\log 2+y\log 3 \)
ganeshie8
  • ganeshie8
let \(n = 2^a3^b\) then the number of integers of form \(2^x2^y\) less than or equal to \(n\) is given by \[\large \sum\limits_{i = 0}^ {\left\lfloor \log_2 n\right\rfloor}\, \left(\left\lfloor\log_3 \frac{n}{2^i}\right\rfloor + 1\right)\]
ganeshie8
  • ganeshie8
that simplifies to \[\large (\left\lfloor \log_2 n\right\rfloor+1)(b+1)+\sum\limits_{i = 0}^ {\left\lfloor \log_2 n\right\rfloor}\, \left\lfloor (a-i)\log_3 2\right\rfloor \]
ganeshie8
  • ganeshie8
in other words, if we sort all the numbers of form \(2^x3^y\) in increasing order, the position of \(n=2^a3^b\) is given by above expression
ganeshie8
  • ganeshie8
Here are some other facts https://oeis.org/A003586 I really want to figure out the solution but I need to go for now, wil try again over the night if it still remains unsolved..
anonymous
  • anonymous
this is called the (generalized) hamming problem
UsukiDoll
  • UsukiDoll
I think the pattern could be predictable because of this theorem Also numbers that are divisible by neither 6k - 1 nor 6k + 1, for all k > 0 like if we let k = 1 then 6(1)-1 = 6-1=5 6(1)+1 = 6+1=7 so maybe that would mean that numbers can't be divisible by 5 nor 7. We don't see 5 10 15..... on that list and similarly we don't see 7 14 21 on that list either. so if k = 2, then 12-1 =11 or 12+1 =13 which means numbers can't be divisible by 11 nor 13. And again, we don't see 11,22,33 nor 13,26,39
anonymous
  • anonymous
@ganeshie8 has the right kind of idea; suppose we had \(2^4 3^7\) -- we could take a look at a few candidates that are bigger but only by somewhat: $$2^5 3^7,2^3 3^8,2^2 3^9,\dots$$ note that increasing the exponent of the 2 is twice as big; if we trade a 2 for a 3, we'll get something 3/2 or one-and-a-half times as big; etc.
anonymous
  • anonymous
we could also trade three 2s for a 3, which is only 9/8 = 1.125 as big; we're interested in making \(3^a,2^b\) as close as possible so that we can make their ratio a small number greater than \(1\) -- examples: $$\frac32=1.5\\\frac98=1.125\\\frac{32}{27}=1.\overline{185}\\\frac{256}{243}\approx1.0535$$and so on
anonymous
  • anonymous
essentially, we're looking for simple rational numbers close to \(\log_2(3)\approx1.5850\)
anonymous
  • anonymous
in terms of logs, we're looking for ways to exchange \(\log2,\log3\) terms in \(x\log 2+y\log 3\) to get something slightly bigger -- in other words, we want to find integers \(a,b\) such that \(a\log 2+b\log 3\) is small and positive
UsukiDoll
  • UsukiDoll
at least I tried :(
ganeshie8
  • ganeshie8
from hamming problem, it seems they are generating the sequence by multiplying a prime to a term already existing in the sequence... it is essentially an algorithm... i think we cannot find a "formula" for the next term...
ganeshie8
  • ganeshie8
i mean, if such a formula exists, then there is no need to use an algorithm to generate the sequence recursively.. (ofcourse cost could be a reason but im beginning to feel that a direct formula is impossible.. )
ganeshie8
  • ganeshie8
Here is a fiddle if you want to play with it more.. https://jsfiddle.net/ganeshie8/uzj8prf6/ @UsukiDoll

Looking for something else?

Not the answer you are looking for? Search for more explanations.