This is another wonderful puzzle from *A Moscow Math Circle, Week by week problem Sets* by Sergey Dorichenko.

There are two parts to the problem:

(a) Place seven stars in a 4×4 grid so that, no matter which two rows and which two columns are erased, at least one star will remain.

(b) Prove that if six stars are placed in a 4×4 grid, one can always erase all of them by erasing two rows and two columns.

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy Eid to everyone!

C and F independently are connected to all except one circle. So, C and F cannot have two adjacent numbers. So C is 1 and F is 8. So A is 7 and H is 2.

3,4,5,6 needs to be placed in B,D,E,G

B and D cannot be 6

E and G cannot be 3

Since the image is symmetric across vertical line, lets say B is 3. If E is 6, and D and G become 4 and 5 which is not possible. So G is 6. So D is 4 and E is 5.

So, A is 7, B is 3, C is 1, D is 4, E is 5, F is 8 and G is 6.

Hope you all enjoyed the puzzle!