maths expert

• starblue

  21 Sep 05, 19:43

  wah kao... math whizes always amazes me. but they scare me also...
• CoolMyth

  21 Sep 05, 20:56

  Maths Olympiad stuff.....
  
  I suspect as much.
• mmmm123

  21 Sep 05, 22:45

  Originally posted by Ray87:

  2)find the smallest three digit number n such that if the three digits are a, b, and c, then n= a + b + c + ab +bc + ac +abc

  Recast it as follows: (where each a,b,c is a positive single digit integer)
  
  n = [a,b,c] = 100a+10b+c = a + b + c + ab + ac+ bc +abc eqn (1)
  
  Since it is asking for the smallest digit possible, let's take a = 1 and the above eqn. reduces to:
  
  99 = 2bc + c - 8b eqn (2)
  
  Since b, c needs to be positive, c >= 4 and for them to be single digit, c has to be:
  
  c = 9
  
  For this allows Eqn (2) to be reduced to the following:
  
  99 = 18b + 9 - 8b
  90 = 10b
  thus: b = 9
  
  Therefore: n = [a,b,c] = 199
  
  I think this ans. is correct. Anyone has another solution technique ??
  
  
• CoolMyth

  21 Sep 05, 23:03

  Okay, I finally understood Q ii).....
  
  Shall be 199 then.
• mmmm123

  22 Sep 05, 00:32

  Originally posted by Ray87:

  1)try this. find the last three digits of 9^100 -1(without the use of calculator)
  
  

  Write as follows:
  
  9^100-1 = ( -x + 10 )^100 -1 ; where x =1
  
  Expand using standard binomial theoream up to the first three terms, we have:
  
  ( -x + 10 )^100
  = (-x)^100 + 100 * (-x)^99 * 10 + 100 * 99/(2) * (-x)^98 * 10^2 + .....
  
  subs. x = 1 into the above, we have:
  
  ( -x + 10)^100 = 1 - 1000 + 495000 + .....
  
  Note: By inspection higher order terms ( > third term ) contain number with at least 4 zeros behind it and gets progressively larger in abslute value and therefore has no effect on the last three digits of the sum of the first two terms )
  
  Thus: the last three digits are no more than '000'
  
  ans. 000
  
  Uisng similar technique, the last four digits can be deduced as '4000'
  
  
• findingnewidea

  22 Sep 05, 08:42

  t = -2.67808
  
  t = -1.10728
  
  t = 0.463516
  
  t = 2.03431
  
  ... (only 4 solutions are given here)
• Ray87

  22 Sep 05, 20:48

  my prelims are finally over.. yay
• Ray87

  22 Sep 05, 20:52

  nice one mmm123, i couldnt have thought of that. i was spinning my pen throughout the olymiad.
• 798

  22 Sep 05, 21:46

  Originally posted by valentinoo:

  4sin2t + 3cos2t = 5cos(2t - 53.1 degrees)
  
  i forget how to do...
  
  neone knows?

  for me... i will express as phasor form and do.
  
  To phasor form:
  4sin2t = 4/sqrt(2) * arg(0deg)
  
  3 cos2t = 3 sin(2t + 90deg) = 3/sqrt(2) * arg(90deg)
  
  4/sqrt(2) * arg(0deg) + 3/sqrt(2) * arg(90deg)
  = 4/sqrt(2) + 3i/sqrt(2)
  = 2.828 + 2.121i
  = 3.5355 arg(36.9deg)
  
  3.5355 arg(36.9deg) converts back to time domain function
  
  3.5355 arg(36.9deg)
  = 3.5355*sqrt(2) sin(2t + 36.9deg)
  = 5 sin(2t + 36.9deg)
  = 5 cos(2t + 36.9deg - 90deg)
  = 5cos(2t - 53.1degrees)
  
  
• Beep

  22 Sep 05, 23:35

  Hey, are you from the RJC Maths Olympiad team?
  
  Yes yes the s01a one with all the chinkish scholars.