3
$\begingroup$

Attached Grid (Total 5 x 5) has a black box grid (3 x 3) inside it.

The Black box grid needs to be filled out with numbers 1 thru 9 using all numbers once.

The Outer Red box squares (Total 16 ) need to be filled out with numbers 10 thru 25 using all numbers once.

All numbers must be used once.

Some numbers are already entered to get started.

The goal is SUM 50.

Rows and Columns 1,2 and 3 (5 numbers in each) must individually add up to 50. So each Row and Column (1,2 and 3) must add up to 50.

Also each diagonal, two as shown, (5 numbers) must add up to 50

Rows X and Y and Columns X and Y can add up to anything, no restrictions.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Downvoting because this doesn't seem to be uniquely solvable. $\endgroup$ – Rand al'Thor Aug 15 '17 at 13:54
  • $\begingroup$ I could only get 1 answer. Let us see how many answers come back. $\endgroup$ – DEEM Aug 15 '17 at 14:02
  • $\begingroup$ There is only 2 answers (as shown by Apep) but that's because so many already given values. If you gave no values in the middle and the outside values are set, then you could still work it out. $\endgroup$ – BMS21 Aug 15 '17 at 15:25
  • $\begingroup$ @Randal'Thor Only One answer given. $\endgroup$ – DEEM Aug 23 '17 at 12:49
3
$\begingroup$

One solution is:

solution

How I approached solving:

First, the inner box can be solved as a regular magic square. Each row, column and diagonal would add up to 15. Columns and rows with 1 empty space can be solved by subtracting the other two numbers from 15.

For the remainder:

The rows, columns, and diagonals of the inner box all sum to 15, so the remaining 2 numbers for rows 1, 2, 3; columns 1, 2, 3; and diagonals should sum to 35 (50 - 15). Simply subtracting the existing number on the outer layer from 35 gives us number on the opposite side. The diagonal from top-left to bottom-right can be filled in with the remaining numbers, 25 & 10, in either order.

$\endgroup$
  • $\begingroup$ How did you know the middle 3x3 was a magic square? $\endgroup$ – Mike Earnest Aug 15 '17 at 15:37
  • $\begingroup$ @Mike I guessed. It should work as long as the outer layer doesn't fill 2 opposite squares with non-complements and only uses a complement pair in opposite squares. In this case, no complement pairs were used and no opposite squares were filled. $\endgroup$ – Apep Aug 15 '17 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.