 
24th November 2011, 10:34 PM

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A little innocent looking maths problem
Find the number ( less than 1000) which leaves a remainder of 1 when divided by 7, a remainder of 2 when divided by 11 and a remainder of 3 when divided by 13. What is the next number (greater than 1000) which also satisfies the above three conditions?
Something for you students to ponder over. Peace.
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25th November 2011, 09:09 AM

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Re: A little innocent looking maths problem
211, 1212  from my smartass friend

 
25th November 2011, 09:44 AM

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Re: A little innocent looking maths problem
Very good. The more important question is, what was the method your smartie pants friend used? I hope its not by trial and error.

 
25th November 2011, 10:19 AM

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Re: A little innocent looking maths problem
Anyways for those who wish to view the detailed solutions, I have done them up and posted it here: http://www.whitegroupmaths.com/2011/...tnumbers.html
Hope it helps. Peace.

 
25th November 2011, 10:28 AM

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Re: A little innocent looking maths problem
i used number theory... lol. Anyway interesting solution, one must have really good observation for that.

 
25th November 2011, 01:45 PM

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Re: A little innocent looking maths problem
Let a be the number. a=7b+1, or 11c+2, or 13d+3, so 7b=11c+1=13d+2, so 7b give remainder 1 when divided by 11.
By some theorem, b gives a remainder of 8, so b8 is divisble by 11. Now 7b=13d+2, so 7b divided by 13 give remainder 2, some theorem later b gives remainder of 4 when divided by 13. So let b4=13e.
Now 13e4 must be divisble by 11(b8 from just now) by some theorem, e gives a remainder of 2 when divided by 11. So e=2, 13, 24,... Sub it back and get a.

 
25th November 2011, 06:38 PM

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Re: A little innocent looking maths problem
Sounds like a complicated Maths problem, to me
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25th November 2011, 08:35 PM

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Re: A little innocent looking maths problem
Originally Posted by Nish Let a be the number. a=7b+1, or 11c+2, or 13d+3, so 7b=11c+1=13d+2, so 7b give remainder 1 when divided by 11.
By some theorem, b gives a remainder of 8, so b8 is divisble by 11. Now 7b=13d+2, so 7b divided by 13 give remainder 2, some theorem later b gives remainder of 4 when divided by 13. So let b4=13e.
Now 13e4 must be divisble by 11(b8 from just now) by some theorem, e gives a remainder of 2 when divided by 11. So e=2, 13, 24,... Sub it back and get a. LOL "some theorem'. Yeah your friend is correct, the theorem is actually called the linear congruence theorem. Peace.

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27th November 2011, 11:26 PM

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Re: A little innocent looking maths problem
Its rather pleasing to see some introductory number theory question here.
It is like our common "Chinese remainder problem" , the general solution to the problem is 1001n+211.
Certainly the only positive solution less than 1000 is 211 and the following solution is 1212.
Whitecorp, it is rather interesting to see a linear algebra approach but it will not be bound to integer solution, which is saddening here.
Last edited by Icystrike; 27th November 2011 at 11:29 PM.

 
27th November 2011, 11:51 PM

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Re: A little innocent looking maths problem
I said goodbye to my maths 3 years ago when I got my "O" level results and went to poly. And now looking at all those numbers hurts my head.

 
28th November 2011, 01:28 AM

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Re: A little innocent looking maths problem
Originally Posted by Icystrike
Whitecorp, it is rather interesting to see a linear algebra approach but it will not be bound to integer solution, which is saddening here. Yeah, which is why I imposed further specific conditions on the unknowns after obtaining the general solution set to do further filtration.

 
28th November 2011, 01:32 AM

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Re: A little innocent looking maths problem
Originally Posted by Aiyilin I said goodbye to my maths 3 years ago when I got my "O" level results and went to poly. And now looking at all those numbers hurts my head. And what are you studying currently?
You can throw away most of the math you learnt, just make sure you know how to add, subtract multiply and divide when dealing with moneyrelated stuff.

 
29th November 2011, 01:27 AM

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Re: A little innocent looking maths problem
Another one here:
s = km
p = k/m
sp = 3
k = ?
Have fun. Peace.

 
1st December 2011, 08:10 PM

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Re: A little innocent looking maths problem
is the answer for the lastest qn like k = m/2 + sqrt[ m(m12) ]/2 for m<0 and m>=12?

 
1st December 2011, 08:45 PM

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Re: A little innocent looking maths problem
Originally Posted by megaman123 is the answer for the lastest qn like k = m/2 + sqrt[ m(m12) ]/2 for m<0 and m>=12? hahahaha please say no. megaman go makan with me leh, gossip again. hahaha

 
1st December 2011, 09:34 PM

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Re: A little innocent looking maths problem
Originally Posted by megaman123 is the answer for the lastest qn like k = m/2 + sqrt[ m(m12) ]/2 for m<0 and m>=12? No the answer is not that crazily complicated.
k=4 and m=16.

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1st December 2011, 09:46 PM

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Re: A little innocent looking maths problem
Originally Posted by whitecorp No the answer is not that crazily complicated.
k=4 and m=16. oh... but my answer still kinda works for any k and m. lol

 
1st December 2011, 09:57 PM

#18 (permalink)
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Re: A little innocent looking maths problem
Originally Posted by megaman123 oh... but my answer still kinda works for any k and m. lol Your answer does work, however they are looking for integer answers for both k and m. But based on your formula, if you choose m to be an integer, your k would be a noninteger; unless of course you pick the magical combination.
Its a great try anywayswell done. Keep it up. Peace.
Last edited by whitecorp; 1st December 2011 at 10:03 PM.

 
1st December 2011, 10:18 PM

#19 (permalink)
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Re: A little innocent looking maths problem
You might have come across this rather popular question:
The 4 pieces appear to have been merely shuffled around, yet in the process a gap gets created.
The reason?

 
2nd December 2011, 01:55 AM

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Re: A little innocent looking maths problem
Originally Posted by whitecorp You might have come across this rather popular question:
The 4 pieces appear to have been merely shuffled around, yet in the process a gap gets created.
The reason? Easy, came across this while wikipediaing long time ago. Missing square puzzle
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