Find three consecutive odd integers

Question

Find three consecutive odd integers such that three times the first and third is no less than five times the second.

Answer 1

Hi Karl. This is a fun one!

Find three consecutive odd integers such that three times the first and third is no less than five times the second.

Let's break it up and translate the mathspeak so we can solve it:

1. three consecutive odd integers = three whole numbers that follow each other and are not divisible by 2 (like 1, 3, 5 or 7, 9, 11)

We can use x, x+2 and x+4 as our numbers

1. three times the first (x) and third (x+4) is no less than five times the second (x+2)

The wording here is a big clue that we are talking about an inequality; no less than means greater than or equal to (≥). Sneaky.

three times the first (x) and third (x+4) = 3(x+x+4)

five times the second (x+2) = 5(x+2)

Now to put that into our inequality and solve:

3(x + x+4) ≥ 5(x+2)

6x+12 ≥ 5x+10

x ≥ -2

But is that right? The only way to find out is to put some real numbers in and see if it works:

Let x=-1, x+2=1 and x+4=3

3(-1+3) ≥ 5(1)

3(2) ≥ 5

6 ≥ 5 ✔

And

Let x=3, x+2=5 and x+4=7

3(3+7) ≥ 5(5)

3(10) ≥ 25

30 ≥ 25 ✔

I think we did it! 😁

Answer 2

Agreed.... so we'll do it twice

1st is x

2nd is x+2

3rd is x+4

"three times the first, but the third is NOT tripled"

3x + x+4 >= 5(x+2)

4x + 4 >= 5x + 10

4 >= x + 10

-6 >= x

since x must be an odd integer, x=-7, -9, -11, etc..

in the best case they are -7, -5, -3

3(-7) + -3 = -21 + -3 = -24 while 5(-5) = -25 and the inequality hold

vs.

"the first and the third are BOTH tripled"

3[x + x+4] >= 5(x+2)

3[2x+4]>=5x + 10

6x + 12 >= 5x + 10

x + 12 >= 10

x >= -2

again since x must be odd integer, then x=-1,1,3,5,....

the worst case is -1,1,3

3(-1 + 3) = 3(2) = 6 while 5(1) = 5 and the inequality holds

We have done our due diligence