My Math Forum please find minimal n

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 December 22nd, 2012, 06:12 AM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 please find minimal n $n>(\sqrt 2 + \sqrt 3)^6$ $\text where \,\,n\in N\text$ $\text please \,\,find \,\, minimal \,\, n\text$ $\text Ans:\,\, 970\text$
 December 22nd, 2012, 08:33 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: please find minimal n Why have you not just done the arithmetic? If you have a calculator this is easy. And apparently you have a computer! If you are running windows, you can use the calculator that comes with it: square root of 2 is about 1.414214 and square root of 3 is about 1.732051 so that $\sqrt{2}+ \sqrt{3}= 3.14626$. That to the 6 th power is a little bit larger than 969 so the smallest integer larger than that is 970.
December 22nd, 2012, 07:11 PM   #3
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Quote:
 HallsofIvy wrote : Why have you not just done the arithmetic? If you have a calculator this is easy. And apparently you have a computer!
sorry! no use of calculator or computer.

 December 22nd, 2012, 07:20 PM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 Re: please find minimal n Wierd that root 2 plus root 3 gives pi correct to 2 decimal places...
 December 22nd, 2012, 08:04 PM #5 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 Re: please find minimal n If we use the binomial theorem we only have 1 difficult calculation, root 6. $(\sqrt{2})^6 + 6 (\sqrt{2})^5 \sqrt{3} + 15 (\sqrt{2})^4 (\sqrt{3})^2 + 20 (\sqrt{2})^3 (\sqrt{3})^3 + 15(\sqrt{2})^2 (\sqrt{3})^4 + 6 \sqrt{2} (\sqrt{3})^5 + (\sqrt{3})^6$ $8 + 24 \sqrt{6} + 180 + 120 \sqrt{6} + 270 + 54 \sqrt{6} + 27$ $485 + 198 \sqrt{6}$ Up to this point, no messy arithmetic involved, just integer addition and integer multiplication. Now, there are many easy ways to extract the root of 6 manually to 5 or 6 decimal places and then multiply by 198, i used 198(2.4494 = 484.997. Since this is close to 485 i rounded the 8 to 9 and used 2.44949 and since i stayed under 485 i knew i was safe... 485 + 484.997 = 969.997 so 970 is the answer.
 December 22nd, 2012, 09:11 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: please find minimal n a=sqrt(2):b=sqrt(3) ((a+b)^3)^2 = (11a + 9b)^2 = 485 + 198ab = 485 + 198SQRT(6) (YEA...as per our favorite Agent!) n = 485 + 198SQRT(6) (n - 485) / 198 = SQRT(6) So , knowing 3 > SQRT(6) > 2, then 1079 > n > 881; so Albert must have a "trick of his own" to wrap that up cleanly... (even if agentredlum's is accurate...but "seems" to need a calculator a bit)
 December 23rd, 2012, 01:06 AM #7 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: please find minimal n and 24 < sqrt(600) < 25 so 2.4 < sqrt(6) < 2.5 Now, (a + 1)^2 - a^2 ~= 2a and 60000 - 57600 = 2400 = 2a*m where a = 240 so m, = 5 and sqrt(60000) ~= 245. 245^2 = 240 * 250 + 25 > 60000 so 244 < sqrt(60000) < 245. Gives 2.44 * 198 = 2.44 * (200 - 2) = 488 - 4.88 = 483.12 2.45 * 198 = (2.44 + 0.01) * 198 = 483.12 + 1.98 = 485.1 Yields 485 + 483.12 = 968.12 < 485 + 198sqrt(6) = (sqrt(2) + sqrt(3))^6 < 485 + 485.1 = 970.1 And n = 969 or 970 without calculator. Since 240 * 250 + 25 = 60025 is slightly more than 60000, one might argue that sqrt(60000) > 244.7 and so 198 * sqrt(6) > 198 * 2.447 = 198 * (2.44 + 0.007) = 483.12 [from above] + (200 - 2) * 0.007 = 483.12 + 1.4 - 0.014 > 484 so n > 969 which implies n = 970
 December 23rd, 2012, 01:14 AM #8 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: please find minimal n keep trying ! I will post a very tricky method ! (not mine) soon
 December 23rd, 2012, 03:02 AM #9 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: please find minimal n using binomial expansion : $(\sqrt 3 + \sqrt 2)^6 +(\sqrt 3 - \sqrt 2)^6=2[27+15(18+12)+8]=970$ $for: \,\,0<(\sqrt 3 - \sqrt 2)^6<1$ we get the minimal n=970
December 23rd, 2012, 06:56 AM   #10
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 Originally Posted by Denis (YEA...as per our favorite Agent!)
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