The sum of 3 consecutive even integers is 408. find the integers

Let n be the unknown value

n + (n+2) + (n+4) = 408

Then we would simplify by combining like terms

n + n + 2 + n + 4 = 408

3n+6=408
     -6     -6

3n=402

Divide. 

3n=402
--    
3      3

n = 134. 
So the first integer is 134. Then plug in the rest to the n+2 and n+4

134+2= 136
134+4= 138

The integers are then 134, 136, 138

How many is in each case if there is 12 in each case it would be 40

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