奧林匹克數學

1)2²º¹¹= (2⁶)³³⁵ x 2= (13 x 5 - 1)³³⁵ x 2因為 (13 x 5 - 1)³³⁵ ÷ 13 的餘數 = - 1³³⁵ = - 1
所以 (13 x 5 - 1)³³⁵ x 2 ÷ 13 的餘數 = - 2即 2²º¹¹ ÷ 13 的餘數 = - 2 + 13 = 11
2)設三個高分別 = 3h , 4h , 5h△ = (1/2) AB * 3h = (1/2) BC * 4h = (1/2) CA * 5h3AB = 4BC = 5CAAB : BC = 4 : 3 = 20 : 15 , BC : CA = 5 : 4 = 15 : 12故 AB : BC : CA = 20 : 15 : 12整數 AB 最小 = 20
3)設長 = L , 闊 = W , 則L W = 2(L + W) + 92L + 2W - L W - 4 + 13 = 0L(2 - W) + 2(W - 2) + 13 = 0(W - 2) (2 - L) = - 13(W - 2) (L - 2) = (1) (13)W - 2 = 1 , L - 2 = 13W = 3 , L = 15周界 = 2(W + L) = 2(3 + 15) = 36
4)x² + y² = 1
y² = 1 - x²2x + 5y²= 2x + 5(1 - x²) = - 5x² + 2x + 5= - 5(x² - (2/5)x - 1)= - 5(x² - 2 * 1/5 * x + 1/25 - 1 - 1/25)= - 5 ( (x - 1/5)² - 26/25 )= - 5 (x - 1/5)² + 26/5當 x = 1/5 , 極大值 = 26/5