Question 1108354
Find the three consecutive even numbers where the one-fourth of the first and the third is equals to the difference between the second digit minus twenty-two

Let the three  even numbers be (n-2) , n , (n+2)

{{{(1/4)((n-2)+(n+2))}}} ={{{ n-22}}}

{{{(1/4) (2n) = n-22}}}

2n = 4n -88

2n = 88

n=44

the numbers are 42,44,46