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 reply to

Numbers involved by Chris, PhD)

Continuing from my previous comments:

So far we have deduced that a = 8 and b = 1.

Also we have deduced that (z1) = 0 and (z6) = 5. That is:

1 _ _ _ _8 +

1_ _ _ _ 8+

1_ _ _ _ 8 +

1_ _ _ _ 8 +

1_ _ _ _ 8 +

1_ _ _ _ 1 +

= ?

It is now a simple exercise to determine the other values of (zi).

We perform addition the usual way, 5*8 +1 = 41. Therefore we write the last digit as 1, and bring a remainder 4 to the next column.

(r4)

1 _ _ _ _8 +

1_ _ _ _ 8+

1_ _ _ _ 8 +

1_ _ _ _ 8 +

1_ _ _ _ 8 +

1_ _ _ _ 1 +

=

_ _ _ _ _ 1

The next digit is thus determined using the formula:

8*(z2) + 1*(6-z2) + any remainders carried over(R)

= 7*(z2) + 6 + R

In this case, R = 4.

Therefore formula becomes,

= 7*(z2) + 10

This formula gives a last digit of 1 or 8 when (z2) is 3, or 4. Giving 31, or 38. In any case, the next remainder is 3.

Next to determine (z3) we apply the same procedure with 3 as the remainder:

7*(z3) + 6 + R = 7*(z3) + 6 + 3 = 7*(z3) + 9.

This only has a last digit of 1 or 8 when (z3) is 6, giving 51.

Therefore (z3) is 6, and the next remainder is 5.

To find (z4) we apply the same procedure:

7*(z4) + 6 + R = 7*(z4) + 6 + 5 = 7*(z4) + 11.

This only has a last digit of 1 or 8 when (z4) is 1, giving 18.

Therefore (z4) is 1 and the next remainder is 1.

To find (z5) we apply the same procedure:

7*(z5) + 6 + R = 7*(z5) + 6 + 1 = 7*(z5) + 7.

This only has a last digit of 1 or 8 when (z5) is 2 or 3, giving 21 or 28. In both cases the next remainder is 2. This is consistent with (z6)=0, as 7*(z6) + 6 + R = 6+R = 6+2 = 8.

Therefore in summary, we have

(z1) = 5

(z2) = 3 or 4

(z3) = 6

(z4) = 1

(z5) = 2 or 3

(z6) = 0