The solution above is WRONG!
Here's the correct solutions, for both inclusive and exclusive of 200 and 600:
Using the digits 0,1,2,3,4,6 and 7, how many 3-digit even numbers between
200 and 600 can be made?
You didn't state whether it was inclusive or exclusive of 200 and 600,
If it's 'exclusive' the smallest even number allowed is 202, and the largest
even number allowed is 576. If 'inclusive' both 200 and 600 will be included.
However, for calculation purposes, it will be more convenient to include
200 and exclude 600. Then we can adjust the count or sum so as to
include or exclude 200 and 600.
So we will first assume that the smallest even number allowed is 200 and
the largest is 476. That is, we are including 200 but excluding 600. We
will adjust later.
The hundreds digit can only be 2,3 or 4.
So that's 3 ways to choose the hundreds digit.
The tens digit can be any of the 7 digits 0,1,2,3,4,6 or 7
So that's 7 ways to choose the tens digit.
The ones digit must be 0,2,4 or 6, to make sure it's even.
So that's 4 ways to choose the ones digit.
So the answer we get here is 3•7•4 = 84 ways.
Now that includes the number 200, which we are excluding,
So we must subtract one from the 84.
Answer: 83, assuming we are not counting 200 or 600.
If we are including them both then the answer would be 2 more than 83, or 85.
----------------------------
What is their sum?
Think of the long column of 84 3-digit numbers to be added. For
convenience, let's INclude 200 but EXclude 600.
200
202
204
206
...
470
472
474
+476
------
sum
The 84 hundreds digits in the leftmost column contain an equal number of the 3
digits 2, 3 and 4. Since 84/3 = 28. the sum of all the hundreds digits is 28(2+3+4) = 252.
Therefore the sum of all hundreds digits contributes 252∙100 or 25200 to
the final sum.
The 84 tens digits in the middle column contain an equal number of all 7 digits
0,1,2,3,4,6 and 7. Since 84/7 = 12. the sum of all the hundreds digits is 12(0+1+2+3+4+6+7) = 276.
Therefore the sum of all tens digits contributes 276∙10 or to the sum.
The 84 ones digits in the rightmost column contain an equal number of the 4
digits 0,2, 4 and 6. Since 84/4 = 21. the sum of all the hundreds digits is
21(0+2+4+6) = 252.
Therefore the sum of all ones digits contributes 252 to the final sum.
Therefore the sum which includes 200 but excludes 600 is 25200+2560+252 = 28212.
(Chances are this is NOT the answer you want, but it is the easiest to calculate
and from which the other two possible answers are easily calculated.]
If we want the sum which excludes both 200 and 600 we must subtract 200 from
28212 and get 28012.
If we want the sum which includes both 200 and 600 we must add 600 and get
28812.
Edwin