Question
Find the unit digit of the expression: 1! + 2! + 3! +
4! + 5! + ........ + 2027!.Solution
1! = 1 2! = 1 X 2 = 2 3! = 1 X 2 X 3 = 6 4! = 1 X 2 X 3 X 4 = 24 5! = 1 X 2 X 3 X 4 X 5 = 120 6! = 1 X 2 X 3 X 4 X 5 X 6 = 720 Here, we can observe that the unit-digit of numbers larger than 4! Is 0. So, unit digit of the given expression, 1! + 2! + 3! + 4! + 5! + ........ + 2027! = unit digit of (1 + 2 + 6 + 4 + 0) = unit digit of 13 Required unit digit = 3
Equation 1: x² - 200x + 9600 = 0
Equation 2: y² - 190y + 9025 = 0
I: x2Â + 31x + 228 = 0
II: y2 + 3y – 108 = 0
I). 5p2 Â - p - 4 = 0
II). q2 - 12q + 27 = 0
I). p2 - 26p + 165 = 0
II). q2 + 8q - 153 = 0
l). 3p + 2q = 27
ll). 4p - 3q = 2
I. 2b2 + 31b + 99 = 0
II. 4a2 + 8a - 45 = 0
I. 6x² + 77x + 121 = 0
II. y² + 9y - 22 = 0
I. 2p2 - 3p – 2 = 0 II. 2q2 - 11q + 15 = 0
I: √(100 x4 + 125x4) + 7x + 41/2 = -4x
II: 3√(64y3) x 2y + 19y + 72 = -3y +...
If the roots of the quadratic equation 6m² + 7m + 8 = 0 are α and β, then what is the value of [(1/α) + (1/β)]?
