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 November 24th, 2012, 11:03 PM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 find max(n) and m $\text m,n \in N\text$ $\text n=\sqrt {m-184} + \sqrt{m+24}\text$ $\text find : max(n) \,\, and \,\, m\text$ Ans :max(n)=104 , m=2785
 November 24th, 2012, 11:13 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: find max(n) and m Seems to me m and n are unbounded.
November 24th, 2012, 11:36 PM   #3
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Re: find max(n) and m

Quote:
 MarkFL wrote :Seems to me m and n are unbounded.
if m and n are positive real numbers then m and n are unbounded

but here m and n are natural numbers (the solutions of m and n are limited), we can find max(n) and corresponding m

 November 24th, 2012, 11:44 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: find max(n) and m D'oh! Yes, my apologies...
 November 25th, 2012, 01:55 AM #5 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: find max(n) and m Let $1) \ m - 184 \= \ k^2$ and $2) \ m + 24 \= \ (k+a)^2$ for natural numbers m, k, a. subtract, 2) - 1) $208 \= \ 2ka + a^2$ we want maximum positive integer k therefore we need minimum positive integer a. try a = 1 208 = 2k + 1 impossible try a = 2 208 = 4k + 4 k = 51 m = 51^2 + 184 = 2785 $n \= \ sqrt{ 2601 } + sqrt{ 2809 } \ = \ 51 + 53 \ = \ 104$
 November 25th, 2012, 02:16 AM #6 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: find max(n) and m excellent
 November 25th, 2012, 03:27 PM #7 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: find max(n) and m now please find : min(n) and corresponding m (m and n are also positive integers) Ans:min(n)=26 , m=265
 November 25th, 2012, 04:55 PM #8 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: find max(n) and m proceed as before with $1) \ 208 \= \ 2ka \ + \ a^2$ This time we want the minimum positive integer k wich means we need the maximum positive integer a. a = 14 already makes 1) greater than 208 and odd integers for a have no chance so try a = 12 208 = 24k + 144 impossible for positive integer k try a = 10 208 = 20k + 100 again impossible for positive integer k try a = 8 208 = 16k + 64 k = 9 m = 9^2 + 184 = 265 $n \= \ \sqrt{265 - 184} \ + \ \sqrt{265 + 24} \ = \ \sqrt{81} \ + \ \sqrt{289} \ = \ 9 \ + \ 17 \ = \ 26$
 November 25th, 2012, 05:29 PM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: find max(n) and m ....and the in-between one is: 24, 28, 52 agentredlum = red gunmetal (anagram)
November 25th, 2012, 06:39 PM   #10
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Joined: Jul 2011
From: North America, 42nd parallel

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Re: find max(n) and m

Quote:
 Originally Posted by Denis ....and the in-between one is: 24, 28, 52 agentredlum = red gunmetal (anagram)
nice, much better than 'ten drug meal'

but i still prefer agent mulder because the character has many qualities i like.

 Tags find, maxn

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