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Old 24th November 2011, 10:34 PM   #1 (permalink)
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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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Old 25th November 2011, 09:09 AM   #2 (permalink)
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Re: A little innocent looking maths problem

211, 1212 - from my smartass friend

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Old 25th November 2011, 09:44 AM   #3 (permalink)
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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.

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Old 25th November 2011, 10:19 AM   #4 (permalink)
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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/...t-numbers.html

Hope it helps. Peace.

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Old 25th November 2011, 10:28 AM   #5 (permalink)
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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.

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Old 25th November 2011, 01:45 PM   #6 (permalink)
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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 b-8 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 b-4=13e.
Now 13e-4 must be divisble by 11(b-8 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.

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Old 25th November 2011, 06:38 PM   #7 (permalink)
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Re: A little innocent looking maths problem

Sounds like a complicated Maths problem, to me

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Old 25th November 2011, 08:35 PM   #8 (permalink)
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Re: A little innocent looking maths problem

Originally Posted by Nish View Post
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 b-8 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 b-4=13e.
Now 13e-4 must be divisble by 11(b-8 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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Old 27th November 2011, 11:26 PM   #9 (permalink)
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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.
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Old 27th November 2011, 11:51 PM   #10 (permalink)
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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.

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Old 28th November 2011, 01:28 AM   #11 (permalink)
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Re: A little innocent looking maths problem

Originally Posted by Icystrike View Post

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.

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Old 28th November 2011, 01:32 AM   #12 (permalink)
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Re: A little innocent looking maths problem

Originally Posted by Aiyilin View Post
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 money-related stuff.

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Old 29th November 2011, 01:27 AM   #13 (permalink)
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Re: A little innocent looking maths problem

Another one here:

s = k-m
p = k/m
sp = -3
k = ?

Have fun. Peace.

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Old 1st December 2011, 08:10 PM   #14 (permalink)
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Re: A little innocent looking maths problem

is the answer for the lastest qn like k = m/2 + sqrt[ m(m-12) ]/2 for m<0 and m>=12?

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Old 1st December 2011, 08:45 PM   #15 (permalink)
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Re: A little innocent looking maths problem

Originally Posted by megaman123 View Post
is the answer for the lastest qn like k = m/2 + sqrt[ m(m-12) ]/2 for m<0 and m>=12?
hahahaha please say no. megaman go makan with me leh, gossip again. hahaha

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Old 1st December 2011, 09:34 PM   #16 (permalink)
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Re: A little innocent looking maths problem

Originally Posted by megaman123 View Post
is the answer for the lastest qn like k = m/2 + sqrt[ m(m-12) ]/2 for m<0 and m>=12?
No the answer is not that crazily complicated.

k=4 and m=16.

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Old 1st December 2011, 09:46 PM   #17 (permalink)
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Re: A little innocent looking maths problem

Originally Posted by whitecorp View Post
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

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Old 1st December 2011, 09:57 PM   #18 (permalink)
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Re: A little innocent looking maths problem

Originally Posted by megaman123 View Post
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 non-integer; unless of course you pick the magical combination.

Its a great try anyways-well done. Keep it up. Peace.


Last edited by whitecorp; 1st December 2011 at 10:03 PM.
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Old 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?

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Old 2nd December 2011, 01:55 AM   #20 (permalink)
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Re: A little innocent looking maths problem

Originally Posted by whitecorp View Post
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 wikipedia-ing long time ago.

Missing square puzzle

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