Jevon Fulbrook brings us this month’s Critical Line challenge, with a puzzling Easter-egg hunt!

Annie the actuary has survived the Christmas party and is now on Easter Break with her family.

She has set up an Easter egg hunt for her daughter and her friends, but was over zealous setting the clues!

The most difficult one – and the clues to find the largest egg, were too hard for the children to solve and they have enlisted your help to solve them.

The location of the egg is reveal by filling in the following by unscrambling a set of letters;

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ !   _ _’_    _ _    _ _ _    _ _ _ _ _ _    _ _ _ _

To solve the location the children will need to unscramble certain letters from the grid of letters given below;

The letters to extract from the grid will be revealed by overlaying the answer found by solving the grid below.

Note that the following 4×4 grid should give you sufficient information to work out how to solve the larger version. Good luck and move quickly, the Easter egg might be melting or will be eaten by something else first!

For your chance to win \$50, send your solution to the puzzle to ActuariesMag@actuaries.asn.au

#### The Critical Line – Volume 11 Solution

The most square numbers found in a magic square by readers was 7, which also matches my own search.

Congratulations to Corey Plover, Stephen Woods, Andrew Parker and Adrian Yiu who all submitted the same solution with a row sum of 541,875

 373^2 360721 205^2 289^2 425^2 527^2 565^2 23^2 222121

The book voucher prize, drawn at random, goes to Stephen Woods.

Further investigating online has led me to conclude that this is the ONLY 7-square 3*3 magic square known to be published (not counting its reflections, rotations and multiples), and that no one has proven or disproven the possibility of an 8-square or 9-square 3*3 magic square.

I should also apologise for my suggestion in the original question that you can freely choose 4 numbers and derive the other 5 when constructing a 3*3 magic square.  In fact you can freely choose only 3 values, provided they are not all on the same one of the 4 lines passing through the centre, and the other 6 are then derived.  And even then, there are some odd/even parity issues that can exclude certain permutations of 3 free numbers.

For the record, here are the 3-square, 4-square, 5-square and 6-square cases with the smallest row sums.

 4 9 2 16 19 4 55 49 16 25 289 121 3 5 7 1 13 25 1 40 79 241 145 49 8 1 6 22 7 10 64 31 25 169 1 265

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