0 x 125 = 0 1 x 125 = 125 2 x 125 = 250 3 x 125 = 375 4 x 125 = 500 5 x 125 = 625 6 x 125 = 750 7 x 125 = 875 8 x 125 = 1000 9 x 125 = 1125 10 x 125 = 1250 11 x 125 = 1375 12 x 125 = 1500 et.seq.,
1 x 125, 5 x 25, 25 x 5, 125 x 1
1 x 250, 2 x 125, 5 x 50, 10 x 25 and 125 + 125 = 250
1 x 125, 5 x 25
Expanded Notation of 125 = (1 x 100) + (2 x 10) + (5 x 1)
0 x 125 = 0 1 x 125 = 125 2 x 125 = 250 3 x 125 = 375 4 x 125 = 500 5 x 125 = 625 6 x 125 = 750 7 x 125 = 875 8 x 125 = 1000 9 x 125 = 1125 10 x 125 = 1250 11 x 125 = 1375 12 x 125 = 1500 et.seq.,
72 = 2 x 2 x 2 x 3 x 3 125 = 5 x 5 x 5 Accordingly, The greatest common factor is: 72 x 125 = 9000
1 x 125, 5 x 25, 25 x 5, 125 x 1
1 x 250, 2 x 125, 5 x 50, 10 x 25 and 125 + 125 = 250
1 x 125, 5 x 25
x^(3) + 125 => x^(3) + 5^(3) This factors to (x + 5)(x^(2) - 5x + 5^(2) ) or (x + 5)(x^(2) - 5x + 25) NB The clue is to spot that the constant (125) is a cubic numbers ( 5^(3)). Similarly with other 'cubic factorings'.
Expanded Notation of 125 = (1 x 100) + (2 x 10) + (5 x 1)
x^(3) + 125 = 0 Remember that 125 = 5^(3) Hence x^(3) + 5^(3) = 0 Factor (x + 5) ( x^(2) - 5x + 25) = 0 So x = -5 & x^(2) - 5x + 125 = 0 Apply Quadratic Eq'n x = {--5 +/- sqrt[)=5)^(2) - 4(1)(125)]} / 2(1) x = { 1 +/- sqrt[25 - 500]} / 2 x = { 1 +/- sqrt[-374] } / 2 This part remain unresolved because you cannot rake the square root of a negative number. Hence x = -5 is the only result.
First 125 numbers sum to 125/2 x 126, so first 125 even numbers sum to double that ie 125 x 126 = 15750
263
1 x 250, 2 x 125, 5 x 50, 10 x 25, 25 x 10, 50 x 5, 125 x 2, 250 x 1 = 250
If 2x2 = 250 then x2 = 250/2 = 125 So x = sqrt(125) = 5*sqrt(5) = +/- 5*2.236 = + or - 11.180