Since sum of the integers along each side is 17, so total of all 3 sides comes to be $17 \times 3 = 51$ , But here each corner is added 2 times. Since sum of all integers from 1 to 9 is 45, so this difference is equal to the sum of three integers at the corners and is equal to $51 - 45 = \boxed 6$

We have ${\color{#D61F06}a}+b+c+{\color{#D61F06}d}+e+f+{\color{#D61F06}g}+h+i=45$ and the sum of sides as $\begin{array}{c} {\color{#D61F06}a}{\color{#D61F06}a}+b+c+{\color{#D61F06}d}+{\color{#D61F06}d}+e+f+{\color{#D61F06}g}+{\color{#D61F06}g}+h+i+{\color{#D61F06}a}=3 \times 17 \\ ({\color{#D61F06}a}+b+c+{\color{#D61F06}d}+e+f+{\color{#D61F06}g}+h+i)+({\color{#D61F06}a}+{\color{#D61F06}d}+{\color{#D61F06}g})=51\end{array} \\ 45+({\color{#D61F06}a}+{\color{#D61F06}d}+{\color{#D61F06}g})=51 \\ {\color{#D61F06}a}+{\color{#D61F06}d}+{\color{#D61F06}g}=6$

Note: Solution edited according to @Lynn Kiaer comment.

Let sum together the three sides so we get:

$2A+B+C+2D+E+F+2G+H+I=17\times3=51=N$

By the other hand the sum of number from $1$ to $9$ is $A+B+C+D+E+F+G+H+I=\frac{(9)(10)}{2}=45=M$

$\therefore$ the sum of the corners is $M-N=51-45=\boxed{6}$

Simple guess and check.

Sum along 3 sides are each 17.

So total is 51

But 3 corners each added two times.

Sum from 1 to 9 is 45

Therefore, the necessary sum is 6

No need to place the numbers. They add to 45 (1+2 +3+...+9). The three sides add to 51 (17 *3) but in that case the corners are double counted. So the answer is 51-45 = 6.

Note that 9,8,7 can't be in the same line, in fact: if 9 and 8 are in the same line 17=9+8+0+0 (impossible); if 9 and 7 are on the same line 17=9+7+1+0 (impossible); if 8 and 7 are on the same line 17=8+7+1+1 or 17=8+7+2+0 (both impossible).

So, 9,8,7 can't be in A,D,G and I have placed them respectively in I,C,F.

Now, note that the line with 9 can be (9,5,2,1) or (9,4,3,1) and the line with 8 can be (8,6,2,1) or (8,5,3,1) or (8,4,3,2)

These two lines must have only a number in common so I choose (9,5,2,1) and (8,4,3,2). The number in common is 2, so I placed it in A. Now, 1 and 5 can be placed in H or G and 3 and 4 can be placed in B or D. Considering the third line (that which contains 7), 17=7+6+3+1, the only combination which contains one of (4,1),(4,5),(3,1),(3,5). So, I placed 1 in G and 3 in D. The remaining three numbers are easy to place.

The answer is $\boxed{6}$

There are only 7 possible combinations of 4 non-duplicated digits selected from 1-9 that add up to 17, and of those, only 3 that contain 3 pairs of duplicates that also include all 9 digits. So it's relatively trivial to enumerate them and visually scan the results for the answer.

this will be:

                                        1

                                 8               9
                                                                                                                        [the sum is exactly 17]
                          6                               4


                   2                  7          5               3

The answer is 6. To add 4 numbers with an odd sum (e.g. 17), you have to either add 3 odd numbers and one even number or 3 even numbers and one odd number. There are 5 odd and 4 even numbers in the first 9 integers, so we need two odds and one even in the corners. Start with the 3 smallest numbers 1, 2, 3 in the corners (they add up to 6), then find the 3 pairs (even, even), (odd, odd) and (even, odd) to make the missing sums of 12, 13 and 14: 5+7=12, 9+4=13, 6+8=14.

So since we need four numbers to equal 17 I split it up into the fact that we need two numbers to make 10 and two numbers to make 7 (because 10+7=17). I started with the possible ways to make 10 and 7 with two numbers:

10:

1+9

2+8

3+7

4+6

7:

1+6

2+5

3+4

Then, I started to make combinations that didn't use the same number twice and only had one matching number from the previous combinations:

1 + 9 + 2 + 5

2 + 8 + 3 + 4 (I knew 2 + 8 + 1 + 6 wouldn't work here because 1 and 2 would overlap with the first combination)

3 + 7 + 1 + 6

Then, I since 1, 2, and 3 were the only numbers that showed up twice in the combinations, I added them up because they were the corners!

1 + 2 + 3 = $\boxed{6}$

Let E represent an even number and O an odd number. Since every side can either have 3 odd numbers and one even number or vice versa, and there have to be 4 even numbers, one of the solutions will look like this:

                  O

              O       E

         O                 O

     E       E       E       O

If we put 2, 4 and 8 in the bottom side (with 2 being in the corner), it works out. We get 1, 2 and 3 in the corners.

Iterate through all possible permutations. That is a matter of second.

View solution...

Just for your information, an example of the solved triangle would look like:

                                          1
                                      5       6
                                   9              7
                                2     8       4      3

Of course, the sum of the three corners are equal to 1 + 2 + 3, which means the three corners add up to 6.

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I worked too hard.

I started with a table of sets of 4 single digit numbers that sum to 17 where a number is used only once per sum.

Since the order of the digits doesn't matter for this table, requiring the digits to be an ascending order eliminates duplication. Since duplicate digits are not allowed, each possible combination can be expressed as an ascending sequence.

1, 2, 5 (since the last can't be more than 9), 9

1, 2, 6, 8

Since the next lower digit in the last place would be 7, making the 3rd place also 7, that is not allowable.

And dropping the 4th below 8 that would result in repeating previously used combinations, so we are done with rows starting 1, 2.

1, 3, 4, 9

1, 3, 5, 8

1, 3, 6, 7

1, 4, 5, 7

We have now enumerated all the possibilities that involve 1.

2, 3, 4, 8

2, 3, 5, 7

There are no more ascending sequences that add up to 17. Since every list of numbers can be arranged in increasing sequence (mumble something about well-ordered sets...) we know that all the possibilities are enumerated.

All of these ascending sequences start with 1 or 2, and all but one of the rows includes 3. This tells us that each combination must include 1 or 2, and all but one include 3. In fact, all the rows except for 1, 4, 5, 7 have two of those three numbers.

So the corners must be 1, 2, 3.

If I had remembered the exact phrasing of the question, I would have realized that I was done. But no, I continued...

I start by placing 1, 2, 3 in the corners. And then I fill one side with the remaining numbers from the first combination: 4 and 9 from 1, 3, 4, 9. I cross off all the other rows that contain a 5 or a 9, because those are no longer available.

I can't use the next row, 1, 2, 6, 8, for the other side that includes 1, because of the 2. So there is only one row left that has a 1, namely 1, 3, 6, 7. I fill the next side in with 6 and 7, and cross off the rows that have 6 or 7 in them.

There is only one row left. And it fits in the remaining side of the triangle.

THE NUMBERS STARTING WITH LEFT CORNER ARE 1 7 6 3 8 4 2 5 9. SO THE REQUIRED SUM IS 6.

• Using simple logic rather than algebra:
• The individual numbers add up to 45
• The 3 sides add up to 51
• The corners are counted twice therefore their sum equates to the difference ie. 51-45 = 6

from the top of triangle clockwise: 1,7,6,3,4,8,2,9,5