la sallle
posted by rr on .
There are 3 consecutive even integers such that the quotient obtained by dividing twice the largest integer by the smallest integer is three less than threefifths of the second integer. What are the integers?

Three consecutive even integers are:
2n, 2n+2, 2n+4, where n is an integer.
The rule is thus:
2(2n+4)/2n = (3/5)(2n+2)  3
Rearranging gives:
0 = 6 n^{2}  19n  20
Find the integer root(s) of this quadratic. Solve for integer n.
Use n to find the three consecutive even numbers. 
I would let my 3 consecutive even integers be
x2, x, and x+2
2(x+2)/(x2) =3x/5  3
expanding and collecting like terms gives us
3x^2  31x + 10 = 0
(x10)(3x  1) = 0
x = 10 or x = 1/3, but x is an integer, so
x = 10
the integers are 8, 10 , and 12