Can someone help me with this: http://assets.openstudy.com/updates/attachments/4e169e120b8bc22757463090-crisis111-1310105170054-scan0004.jpg
please!
lol first step in finding the boxes with presents is to go down the line and open every box ...
I get what they are i trying to ask i just dont know where and how to start
the most basic is to just follow the instructions ... provided they make any sense
Is it even possible to make a chart using whats given?
open all boxes close the evens undo the 3s 3c,6o,9c,12o,15c,18o,21c,24o,27c,30o,33c,36o,39c,42o,45c,48o
what a cheap friends. 1st all fifty box open 2) 2,4,6,8,.....,50th box closed 3) All multiple of 3 except for even numbered box closed.
LOL at imranmeah91
3c,6o,9c,12o,15c,18o,21c,24o,27c,30o,33c,36o,39c,42o,45c,48o 4o,8o,12c,16o,20o,24c,28o,32o,36c,40o,44o,48c 5c,10o,15o,20c,25c,30c,35c,40c,45o,50o
oh! now i see what you are doing! :)
however i dont seem to see a pattern,
occocooocoococooooocccoccccoocooccccocccooooococoo
me either yet :)
wow this is defiantly going to take me some time.
are we spose to keep following this pattern for all 50 boxes?
she went thru it 50 times continuing the pattern
the numbers open at the end are the perfect squares. So there'll be 6 for 50, and 14 for 200.
Yes, i am using my best guess that the pattern if for all 50 boxes/
Trying to get computer to do this
how did you get to that makaze?
The reason that it's the perfect squares it the following: Every time you pass through another number you change the multiples of that number. So a number will be open if it has an odd number of factors, and closed for even. Factors are pairs which multiply together to get the desired number so all of them match and so the number of factors is even. The exception is when a number multiplies with itself to give the number, this is the only time that it has odd number of factors, and so is open.
occoccoocoooocooococccoocccooooocccoocccocooocoooo occocccocoooooooocococooccccooooccooocccooooocooco occocccccoooooocococococccccooocccoooccoooooococco occoccccooooooocooocococccocooocccococcoooooccocco occoccccocooooocooooococccocococccococccooooccoccc ; 10 spots so far
id go with makaze on this one lol
so in the end it would be 6 presents for 50, and 14 for 100??
14 for 200 isn't it?
oh yes my mistake 200.
{1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, \ 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, \ 1, 0, 1, 0} This is what I got
what does the 1 and 0 represent?
and btw thanks a lot for helping me much appreciated!!
1 mean open 0 means closed
oh now i see the pattern in your answer!
Maybe easier way to see it is to think of the prime factorisation. The number of factors for a number n is the product of all the exponents in its factorisation plus 1. So if n=p^aq^b then the number of factors are (a+1)(b+1). So if you want an even number of factors all of the exponents have to be even (as soon as one isn't there'll be a plus 1 at the end) so if you can write n=(p1)^(2a1)...p(q)^(2aq) then n^2=(p1)^(a1)...p(q)^(aq). This was a really good question though :D
so using your method the final answer is 6 for 50 and 14 for 200?
Yes
thank you so much!
<html> <body> <div id="showme"></div> </body> <script language=javascript> var flag=0 var present=new Array() var n,m for (n=1;n<51;n=n+1) {present[n] = "open"} function dothis() {for (m=2;m<51;m=m+1) { for (n=m;n<51;n=n+m) { if (present[n]=="closed"){present[n]="open"} else {present[n]="closed"} } } var show="" for (n=1;n<51;n=n+1) {if (present[n].charAt(0)=="o") {show=show+present[n]+" "+n+" ,"} } document.getElementById("showme").innerHTML = show } dothis() </script> </html> I get :) open 1 ,open 4 ,open 9 ,open 16 ,open 25 ,open 36 ,open 49 ,
course I could be wrong :)
it matches makazes and imrans :)
so the boxes that contain presents are the 1st,2nd,9th,16th,25th,36th,and 49th?
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