Question

The smallest of three consecutive even integers is less than 5 times the largest. Find the integers.

Answer 1

Any ideas?

Answer 2

Let \(2n\) be the first (smallest) integer. Then the other two would be \(2n+2\) and \(2n+4\).

You have \(2n<5(2n+4)\). Know how to solve that?

Answer 3

no

Answer 4

10n+20??

Answer 5

Right, that's what the right side is equivalent to, leaving you with
\[2n<10n+20\\
8n>-20\\~~~~~~\vdots\]

Answer 6

-20 doesnt divide by 8????

Answer 7

Try dividing by 4.

Answer 8

-10/4??

Answer 9

-20/8 simplified to -10/4??

Answer 10

Dividing *both sides of the inequality* by 4, I mean. This gives you
\[2n>-5\]
According to this, the smallest integer can be anything greater than -5. Let me see if this checks out...

Answer 11

-5, -7, -9??

Answer 12

It's not those, no. They have to be even.

Answer 13

-2,-4,-6?

Answer 14

Like I said, any consecutive even integers should work. The closest one to -5 that's greater than -5 is \(2n=-4\). Substituting into the inequality:
\[-4<5(-4+4)\\
-4<0\]
which is true. You'll find that using greater values of \(2n\) will also work.

-2,-4, and -6 do not work, since -6 < -5.

Answer 15

so the 3 are not -4, -2,0??

Answer 16

4, 6, 8??

Answer 17

They definitely can be -4, -2, and 0. (The smallest consecutive evens that make the inequality work.)

I'm saying there are multiple solutions to this problem (since it involves an inequality).

Answer 18

ok I was a little confused but now I got it

Answer 19

let 2x,2(x+1),2(x+2) be the even integers and 2x be smallest,then 2x<5*2(x+2)
2x<10x+20
x<5x+10
-4x<10
x>-5/2
so smallest possible value of x=-2
nos. are-4,-2,0