Posted by **rr** on Monday, September 30, 2013 at 10:42am.

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 three-fifths of the second integer. What are the integers?

- la sallle -
**Graham**, Monday, September 30, 2013 at 12:10pm
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.

- la sallle -
**Reiny**, Monday, September 30, 2013 at 1:27pm
I would let my 3 consecutive even integers be

x-2, x, and x+2

2(x+2)/(x-2) =3x/5 - 3

expanding and collecting like terms gives us

3x^2 - 31x + 10 = 0

(x-10)(3x - 1) = 0

x = 10 or x = 1/3, but x is an integer, so

x = 10

the integers are 8, 10 , and 12

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