# Damaged 4x4 grid

John had a great party yesterday ; but he had a very important piece of paper in his pocket, and an unidentified liquid damaged it. One third of the information is lost. It looks like this now : The yellow numbers are ranging from 2 to 9 (every yellow number is different from all the others). They are assigned to the row/column under them/on their right.

This grid is very important for John. Unfortunately he doesn't remember the pattern he used to create it (John really can't hold his liquor).

Can you help him find the 8 missing numbers ?

John's memory will recover over time, and hints might pop after a few days.

Good luck.

• "This is a 4x4 grid with every number from 2 to 9 assigned to a row/column." Is this right or am I missing something? – Stratos Apr 10 '20 at 22:03
• I edited the image, things about these yellow numbers should be clearer now. – classicalMpk Apr 10 '20 at 22:16

## 1 Answer

If n is the handle (yellow number) for a given row or column then the sum of the numbers in that row/column should be equal to n*(n+1) +1

For example, in the third row, sum = 8*9+1 = 73 • n*(n+1) +1 ... which is the $n$th triangular number $T_n$ plus one. Well spotted. – Rand al'Thor Apr 11 '20 at 0:54
• Exactly ! More precisely, the sum of the numbers is 111 base n (which is n^2+n+1).The hardest part was figuring out the 5, because you have to realize 21+31+43+73=7+13+57+91. Well done. – classicalMpk Apr 11 '20 at 10:26
• Thank you @classicalMpk! Really enjoyed solving this one. Yes, figuring out between 5 and 7 is the tricky part. – Vijay Apr 11 '20 at 10:50