SOLUTION: Find three consecutive multiples of seven such that the product of the largest two exceeds the product of the smallest two by one thousand and seventy-eight

Question 662517: Find three consecutive multiples of seven such that the product of the largest two exceeds the product of the smallest two by one thousand and seventy-eight
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find three consecutive multiples of seven such that the product of the largest two exceeds the product of the smallest two by one thousand and seventy-eight
:
Let x = the middle multiple of 7
then
(x-7), x, (x+7) are the three consecutive multiples of 7
:
" the product of the largest two exceeds the product of the smallest two by one thousand and seventy-eight
:
x(x+7) - x(x-7) = 1078
(x^2 + 7x) - (x^2-7x) = 1078
combine like terms, after removing the brackets
x^2 - x^2 + 7x + 7x = 1078
14x = 1078
x = 1078/14
x = 77 is the middle multiple
:
70, 77, 84 are the three multiples
:
:
see if that flies
(84*77) - 70*77) = 1078
6468 - 5390 = 1078
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