Cosmo has arranged the integers $1,2,\ldots,49$ in a $7\times7$ table, so that in each row (left to right) and in each column (top to bottom) the entries occur in increasing order.

What is the smallest possible sum of the numbers in the fourth row in such an arrangement?
What is the largest possible sum of the numbers in the fourth row in such an arrangement?


1 Answer 1


Assume a table labelled with columns A-G and rows 1-7

Question A : Smallest total

Row 4, Column A has 3 cells above it in the same column of the table and all of these must be smaller than cell 4A, so the smallest it can be is $4$.
Row 4, Column B has 3 cells directly above it, along with cell 4A and all the cells directly above that, making a total of seven cells which must be smaller than 4B, so the smallest it can be is $8$,
Continuing across in this fashion, the lowest total possible is $4+8+12+16+20+24+28 = 112$

Question B : Largest total

Reversing the approach above, the largest value possible in Row 4, Column G is $46$ as there need to be 3 numbers larger below it in the grid.
Row 4, Column F needs 7 number bigger than it, so it can be at most $42$.
Continuing gives $22+26+30+34+38+42+46=238$


![enter image description here]1

  • $\begingroup$ I'd just finished the 'smallest' bit when you posted. Ah well... Your addition's slightly off, though - it's 112, not 116 $\endgroup$ Commented Oct 9, 2015 at 14:29
  • $\begingroup$ @LogicianWithAHat Thanks for the correction, I had typed 28 twice into my addition! $\endgroup$
    – Gordon K
    Commented Oct 9, 2015 at 14:31
  • $\begingroup$ Can you provide a more rigorous proof ? I mean, intuitively, I can see why your answers make sense. But I feel that it still needs to be proven more rigorously that these are optimal . $\endgroup$ Commented Apr 20, 2022 at 2:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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