I need help with this one too, find three integers such that three times?????

let the odd integers be represented by x, y, and z.

1. x+2=y (because they are consecutive odd integers)

2. y+2=z

3. 3x < y+z

3 equations and 3 unknowns, we can easily solve this:

plugging eqn 1 into eqn 2 we get:

x+4=z

plugging this new eqn and eqn 1 into eqn 3 we get:

3x < 2x + 6

3x-2x-6 < 0

x-6 < 0

x < 6

so any odd integer less than 6 will work for the first one.

these combinations are solutions:

1,3,5

3,5,7

5,7,9

You have to assign the three numbers like this:

x is the first number

Then the next consecutive odd number is 2 greater than this: x + 2

Then the next is 2 greater than that: (x + 2) + 2 = x + 4

So, your numbers are:x, x + 2, & x + 4

3 times the first one is just "3x".

This ie 1 LESS than the sum of the other two numbers.....the sum of the other two numbers is: (x + 2) + (x + 4)

==> x + 2 + x + 4

==> x + x + 2 + 4

==> 2x + 6

One LESS than this is: (2x + 6) - 1 which gives 2x + 5

So, you end up with: 3x < 2x + 5

Now, solve for 'x'.

3x - 2x < 5

x < 5

Ok....so x can NOT be EQUAL to 5, it has to be LESS than 5, and it still has to be ODD, so x can either be 3, or 1. This is stated in the problem, and I see some others have overlooked this.

So, if x = 1 then we have for the two other odd numbers:

x + 2 = (1) + 2 = 3

-and-

x + 4 = (1) + 4 = 5

So we have, for the three numbers:

1) 1

2) 3

3) 5

-OR-

If x = 3, then we repeat what 'we' just did and get:

x = 3

==> x + 2 = (3) + 2 = 5

==> x + 4 = (3) + 4 = 7

1) 3

2) 5

3) 7

Either one of these groups of 'answers' satisfies the criteria in the problem and are valid.

I hope this helps, and is clear for you. :)

Let x be the first one.

3x < x + 2 + x + 4

3x < 2x + 6

x < 6

x can be 1, 3 or 5, so your answers are:

{1, 3, 5}

{3, 5, 7}

{5, 7, 9}

I believe these sets are infinite if you extend into the negative integer realm.

let the nos be x x+2 and x+4

3x=(x+2) + (x+4) -1

x=5

the nos are 5 7 and 9

x

x+2

x+4

3(x) -1 = (x+2) + (x+4)

3x-1 = 2x+6

x = 7 and

3 integers are 7, 9, 11