Question

find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the largest number​

Answer 1

A generic odd number can be written as

$2k+1,\quad k \in \mathbb{Z}$

Since there is an odd number every two numbers, three consecutive odd numbers will be

$2k+1,\quad 2k+3,\quad 2k+5$

Now let's make up the equations: the sum of the first two is

$(2k+1)+(2k+3)$

And 27 less than 3 times the largest is

$3(2k+5)-27$

These two must be the same, so we have

$(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3$

Subtracting 4k and 3 from both sides gives

$1=2k \iff k=\dfrac{1}{2}$

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.