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December 22nd, 2012, 06:12 AM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  please find minimal n 
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 . 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  
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: please find minimal n Quote:
 
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. 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 : we get the minimal n=970 
December 23rd, 2012, 06:56 AM  #10  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234  Re: please find minimal n Quote:
 

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