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:facepalm: I don't know what kind of chicks you know. I always go for the bad ass musicians or funny guys.

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I hope Bryce will respond soon, I want to know if I'm right.

 

Also, I was thinking about the problem I wrote last night, and I THINK you can also prove for 121212...1212, after you prove for 1212...12000...00 but I think nobody solved that anyway...

i have no idea what the answer is but its not 25. try substituting it into the original equation.

Wait.

 

Did you mean this:

 

N/d=a+7

(N*5)/d=b+10

 

?

Ok.

 

Then how do you expect me to substitute it, if I know only d? :P

have no idea. I got these out of a booklet i did last year. It didn't come with the answers because it was a competition./

Uhhhh. Why? I don't know much about it but I don't really see the connection.

 

This is what I did:

 

N≡7 (mod d)

N*5≡10 (mod d)

------------------------------

N≡7 (mod d) /*5

N*5≡35 (mod d)

------------------------------

10≡35 (mod d)

 

 

And then I guessed that d could be 25. :uhoh:

And yeah, sorry, I wrote it wrong a few posts back.

:stunned:

haha, i don't even want to show what i did :rolleyes:

maybe it is. i have dont know. it just said to use the euclidean algorithm in the booklet i got with it. anyway, another question:

 

Y1 = 3

Y2 = 14

Y5 = 33

 

Find Y5.

Find the rule for Yn.

I just don't see the connection between that algorithm and the problem...

 

And did you make a typo there? :thinking:

hello guys :hug:

you're so smart. not sure if i can do it, but i'm going to try :rolleyes:

Hi Anita. :nice:

Y1, y1, y3 aren't multiples they are like the number of the terms. so Y1 is the first term, Y2 second etc

my brain hurts, so the first, who knows how long we will be around, and my kids would get the money if i got it and something happend to me. dont know if that would be the same if there was the payout

I'm here.

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