There are six different 6-digit positive integers that add up to a seventh 6-digit integer. Interestingly, all seven of these numbers consist of combinations of only two different digits. That is, only two different digits are used to write the complete set of seven numbers--the same two digits in each number.

So far you can't deduce what the numbers are, but if I were to tell you that seventh number, that is, the total, you'd know what the other six numbers were that made up that total.

What are the seven numbers?

From Enigma No. 1702, "All the sixes",by Ian Kay, New Scientist, 16 June 2012, page 32.

In my previous post (my third post) I promised to explain how I arrived at my solution.

I arrived at my solution by continuing progress from my second post.  The results of my second post were that:
(z1) = 5
(z2) = 3 or 4
(z3) = 6
(z4) = 1
(z5) = 2 or 3
(z6) = 0

where as per before zi refers to the number of 8's in the column when the problem is set up in this format:

_ _ _ _ _ _ +
_ _ _ _ _ _ +
_ _ _ _ _ _ +
_ _ _ _ _ _ +
_ _ _ _ _ _ +
_ _ _ _ _ _

Therefore can immediately deduce the following structure:
1 _ _ 8 _ 8 +
1 _ _ 8 _ 8 +
1 _ _ 8 _ 8 +
1 _ _ 8 _ 8 +
1 _ _ 8 _ 8 +
1 _ _ 8 _ 1

We also note that (z4) = 1, therefore in the 4th column from the right, there exists only one 8.  There are two distinct positions so far where we could place this 8.

 Case A                             Case B
1 _ 8 8 _ 8 +           or     1 _ 1 8 _ 8 +
1 _ 1 8 _ 8 +                   1 _ 1 8 _ 8 +
1 _ 1 8 _ 8 +                   1 _ 1 8 _ 8 +
1 _ 1 8 _ 8 +                   1 _ 1 8 _ 8 +
1 _ 1 8 _ 8 +                   1 _ 1 8 _ 8 +
1 _ 1 8 _ 1 +                   1 _ 8 8 _ 1 +

I will now show why Case B is impossible.  In case B, we so far have 5 numbers of the same structure "1 _ 1 8 _ 8 + ", however there are only 2 blank spaces.  There are only 4 distinct ways of filling these blank spaces, but there are 5 numbers of that structure which must be different (see Charlie's post).  Therefore Case B is impossible, and we take Case A.

Therefore we have deduced the structure so far is:
1 _ 8 8 _ 8
1 _ 1 8 _ 8
1 _ 1 8 _ 8
1 _ 1 8 _ 8
1 _ 1 8 _ 8
1 _ 1 8 _ 1

Now we have four numbers in the structure of "1 _ 1 8 _ 8 + " and 4 distinct ways of filling up the spaces.  That is, (1,8),(1,1),(8,1) and (8,1).  We now do this to arrive at:

1 _ 8 8 _ 8
1 1 1 8 1 8
1 1 1 8 8 8
1 8 1 8 1 8
1 8 1 8 8 8
1 _ 1 8 _ 1

We now have 4 spaces left to fill.

Let us consider the second column from the right.
We know that (z2) = 3 or 4.  If (z2) is 3, then we fill up column two with a 1 and an 8.  But given a final solution to this sum of numbers, we would never be able to distinguish whether we should put 1 at the top and 8 at the bottom, or 8 at the bottom and 1 at the top.  Therefore we deduce that (z2) = 4.  In which case both spaces must be 8.

Now we have:
1 _ 8 8 8 8
1 1 1 8 1 8
1 1 1 8 8 8
1 8 1 8 1 8
1 8 1 8 8 8
1 _ 1 8 8 1

Finally we know that (z5) = 2 or 3.
If (z5) = 3, then we would require to fill one space with an 8 and the other with a 1.  But given the final solution to the sum, we wouldn't be able to figure out which order to place them in.

If (z5) = 2 however, we just fill up both spaces with a 1, and thus given the final sum we would know exactly how to place the numbers.

Therefore the solution is:
1 1 8 8 8 8
1 1 1 8 1 8
1 1 1 8 8 8
1 8 1 8 1 8
1 8 1 8 8 8
1 1 1 8 8 1

This sums up to 818181.

Edited on August 13, 2012, 6:22 am

Posted by Chris, PhD on 2012-08-13 05:36:06