<PRE><B>Question 1130162</B>
&lt;br&gt;
You didn&#39;t say whether digits could be repeated.  I guessed that they could not.  Of course, if they can be repeated, then my response is of no use to you.&lt;br&gt;
Assuming digits can&#39;t be repeated....&lt;br&gt;
(1) For the 3-digit number to be between 200 and 600, the first digit must be 2, 3, or 4.&lt;br&gt;
(2) To be even, the last digit must be 2, 4, or 6; and it can&#39;t be the same as the first digit.&lt;br&gt;
Together, those requirements give us the following list of 3-digit patterns, where the middle digit &quot;_&quot; can be any one of the remaining 5 digits:&lt;br&gt;
2_4
2_6
3_2
3_4
3_6
4_2
4_6&lt;br&gt;
That is 7 patterns, with 5 different values for the middle digit in each pattern; that makes a total of 35 3-digit numbers that can be made.&lt;br&gt;
ANSWER to the first question: there are 35 3-digit numbers that can be made.&lt;br&gt;
An analysis can be made to find the sum of those 35 numbers without simply listing them all out and adding them... but the straight addition might be faster and easier.&lt;br&gt;
But let&#39;s think about how we can find the sum of all 35 numbers by looking at the sum of the digits in each column.&lt;br&gt;
The first (hundreds) digit is 2 10 times, 3 15 times, and 4 10 times.  The sum of the hundreds digits of all 35 numbers is then&lt;br&gt;
(20+45+40)*100 = 10500&lt;br&gt;
The digits in the ones column are 2 10 times, 4 10 times, and 6 15 times.  The sum of the units digits of all 35 numbers is then&lt;br&gt;
(20+40+90)*1 = 150&lt;br&gt;
For the digits in the tens column, consider each of the 7 digits and see how many of the 7 patterns can have that digit as the middle digit.&lt;br&gt;
0 can be used in all 7 patterns
1 can be used in all 7 patterns
2 can be used in 3 patterns
3 can be used in 4 patterns
4 can be used in 3 patterns
6 can be used in 4 patterns
7 can be used in all 7 patterns&lt;br&gt;
The sum of the digits in the tens column of all 35 number is&lt;br&gt;
((7*0)+(7*1)+(3*2)+(4*3)+(3*4)+(4*6)+(7*7))*10 = (0+7+6+12+12+24+49)*10 = 110*10 = 1100.&lt;br&gt;
So the sum of all 35 of the numbers is&lt;br&gt;
10500+1100+150 = 11750&lt;br&gt;
ANSWER to the second question: the sum of the 35 numbers is 11750.</PRE>