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If the questions in that paper had been posed more recently, it would be classified as an easy paper, excluding this years paper of course (as that was pitiful).

Though that may be largely due to the level preparation done by BMO candidates these days compared to a decade ago.
Original post by HLN_Radium
lmfao 11 is the answer that's bullsh*t


U wasteman.


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Original post by physicsmaths


Haha a maths thread has devolved into this.

I thought I would never live to witness this.

:biggrin:
Original post by Xin Xang
Haha a maths thread has devolved into this.

I thought I would never live to witness this.

:biggrin:


First he thought my answer was wrong. Now he corrected himself. What a wasteman fam.


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Original post by physicsmaths
First he thought my answer was wrong. Now he corrected himself. What a wasteman fam.
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yh lol
do you know how answer to Q 5 would go? http://www.bmoc.maths.org/home/bmo1-2012.pdf
Original post by MathMeister
yh lol
do you know how answer to Q 5 would go? http://www.bmoc.maths.org/home/bmo1-2012.pdf


Probably have to look at the multiples and different cases as such.


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Original post by MathMeister
yh lol
do you know how answer to Q 5 would go? http://www.bmoc.maths.org/home/bmo1-2012.pdf


Basically if we prove they cant share a common factor then i think we are done.


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Original post by MathMeister
yh lol
do you know how answer to Q 5 would go? http://www.bmoc.maths.org/home/bmo1-2012.pdf


Cracked it. Got a nice re written neat solution.
ImageUploadedByStudent Room1422304888.414805.jpg
Ignore the let k= bit and k>=5 bit. The square is the key part then the next square after m(m+1)+1 and square before that.
You could go from the perfect squre part and prove that there are no solutions for m(m+1)+1=k^2 but thats long and not needed.



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