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Mathematics 17 Online
OpenStudy (anonymous):

Can someone help me with this: http://assets.openstudy.com/updates/attachments/4e169e120b8bc22757463090-crisis111-1310105170054-scan0004.jpg

OpenStudy (anonymous):

please!

OpenStudy (amistre64):

lol first step in finding the boxes with presents is to go down the line and open every box ...

OpenStudy (anonymous):

I get what they are i trying to ask i just dont know where and how to start

OpenStudy (amistre64):

the most basic is to just follow the instructions ... provided they make any sense

OpenStudy (anonymous):

Is it even possible to make a chart using whats given?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

LOL at imranmeah91

OpenStudy (amistre64):

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

OpenStudy (anonymous):

oh! now i see what you are doing! :)

OpenStudy (anonymous):

however i dont seem to see a pattern,

OpenStudy (amistre64):

occocooocoococooooocccoccccoocooccccocccooooococoo

OpenStudy (amistre64):

me either yet :)

OpenStudy (anonymous):

wow this is defiantly going to take me some time.

OpenStudy (amistre64):

are we spose to keep following this pattern for all 50 boxes?

OpenStudy (amistre64):

she went thru it 50 times continuing the pattern

OpenStudy (anonymous):

the numbers open at the end are the perfect squares. So there'll be 6 for 50, and 14 for 200.

OpenStudy (anonymous):

Yes, i am using my best guess that the pattern if for all 50 boxes/

OpenStudy (anonymous):

Trying to get computer to do this

OpenStudy (anonymous):

how did you get to that makaze?

OpenStudy (anonymous):

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.

OpenStudy (amistre64):

occoccoocoooocooococccoocccooooocccoocccocooocoooo occocccocoooooooocococooccccooooccooocccooooocooco occocccccoooooocococococccccooocccoooccoooooococco occoccccooooooocooocococccocooocccococcoooooccocco occoccccocooooocooooococccocococccococccooooccoccc ; 10 spots so far

OpenStudy (amistre64):

id go with makaze on this one lol

OpenStudy (anonymous):

so in the end it would be 6 presents for 50, and 14 for 100??

OpenStudy (anonymous):

14 for 200 isn't it?

OpenStudy (anonymous):

oh yes my mistake 200.

OpenStudy (anonymous):

{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

OpenStudy (anonymous):

what does the 1 and 0 represent?

OpenStudy (anonymous):

and btw thanks a lot for helping me much appreciated!!

OpenStudy (anonymous):

1 mean open 0 means closed

OpenStudy (anonymous):

oh now i see the pattern in your answer!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so using your method the final answer is 6 for 50 and 14 for 200?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

OpenStudy (anonymous):

thank you so much!

OpenStudy (amistre64):

<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 ,

OpenStudy (amistre64):

course I could be wrong :)

OpenStudy (amistre64):

it matches makazes and imrans :)

OpenStudy (anonymous):

so the boxes that contain presents are the 1st,2nd,9th,16th,25th,36th,and 49th?

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