Question

What are the three consecutive odd integers with a sum of 117?

Answer 1

`38, 39,40`

Explanation:

Let's call the three conscutive numbers `n, n+1, n+2`. Their sum is
`n+(n+1)+(n+2)=117`. This is a linear integer equation.

It can be rewritten as `3n=117-3` from which we deduce that `n=114/3=38`

Answer 2

`37` `39` `41`

Explanation:

Generally we can express an odd integer as follows

`2k-1` where `k` is an integer greater than zero

The next two consecutive odd integers would be

`2(k+1)-1`

`2(k+2)-1`

Proceed as follows

`(2k-1)+[2(k+1)-1]+[2(k+2)-1]=117`

`2k-1+[2k+2-1]+[2k+4-1]=117`

`2k-1+2k+1+2k+3=117`

`6k+3=117`

`6k=114`

`k=114/6`

`k=19`

Now

`2(19)-1=38-1=37`

`2(19+1)-1=40-1=39`

`2(19+2)-1=42-1=41`

`37+39+41=117`

Answer 3

The three number are `{37, 39, 41}`

Explanation:

`color(blue)("Method 1 - Inefficient approach")`

Let `n` be any number. I will call this the seed value

Then by definition `2n` is even. That is, it is divisible by 2.

So `2n+1` is odd

The next number `2n+1+1` will be even so the next one again `2n+1+1+1 ->(2n+1)+2` will be odd.

So every second number will be odd.

Let the first number be `2n+1`

Let the second number be `(2n+1)+2`

Let the third number be `(2n+1)+4`

So we now have

`(2n+1)" "+" "(2n+1)+2" "+" "(2n+1)+4" "=" "117`

`=>6n+9=117`

`=>6n=108`

`n=108/6` but this is the seed value.

So the first value is `2n+1" "->" "2xx108/6+1 = 37`

So the numbers are `{37, 39, 41}`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Method 2 - Efficient approach")`

Let the middle number be `n`

Thus the first number is `n-2`

Thus the last number is `n+2`

The sum of these numbers is:

`(n-2)+n+(n+2) = 117`

`(n-cancel(2))+n+(n+ cancel(2)) = 117`

`3n=117" "color(brown)(larr"Note that "n" is the mean value")`

`n=117/3 = 39`

Thus the three number are `{37, 39, 41}`