Brain Teaser

Question

Place the ten integers 2 through 11 into the ten squares such that the sum of the numbers in each 2x2 square is the same value, P.

What is the maximum possible value of P?


Answer

Each section contains 4 numbers which add up to P, so the sum of all 12 numbers is equal to 3×P.

As two of the squares are used by multiple sections, if we want to maximise P, we need to maximise the duplicate numbers.

However, as P must be an integer, the sum of all squares must be a multiple of 3. This means the highest possible pair of numbers to go in these squares must add to 19, since any higher would make the sum not a multiple of 3. With this, we can now work out our maximum:

3×P=(2+3+4+5+6+7+8+9+10+11)+19=84
P=28


  • Problem by Eric Chen
  • Solution by Binky Mh
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