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A miniature Sudoku puzzle with a twist Question
Fill all 16 squares with a number so that every row, every column and every 2x2 box contains all the numbers 1, 2, 3 and 4 each exactly once. Also the sum of the numbers in the six squares marked with a dot must be equal to the sum of the numbers in the ten squares that are NOT marked with a dot.
1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| will.octagon.gibson | (no comment) | Apr 11, 2026 at 19:04 |
Despite there being no numbers to start with, the extra rule provides a satisfying succession of conclusions that leads to a solution.
Spoilers are hidden in case you want to try this yourself.
Initial insight
The numbers in the finished grid will be the numbers 1 to 4, each repeated 4 times. The numbers 1, 2, 3, 4 sum to 10, so the numbers in the grid will sum to 40. This means that both the set of marked squares and the set of unmarked squares will sum to 20.
Taking the largest possible values for each of the marked squares gives a 4 in the upper left quadrant, 2, 3, 4 (in some order) in the lower left quadrant, and 3, 4 (in some order) in the lower right quadrant. These sum to 20, and changing any of the numbers would result in a smaller sum, so these are the only possible numbers.
Finished solution
1 3 2 4
4 2 1 3
3 1 4 2
2 4 3 1
Solving steps
The initial insight puts a 4 in the only marked square in the upper left quadrant:
# # # #
4 # # #
# # # #
# # # #
This leaves only one marked square where the 4 can go in the lower left quadrant:
# # # #
4 # # #
# # # #
# 4 # #
This in turn leaves only one marked square where the 4 can go in the lower right quadrant:
# # # #
4 # # #
# # 4 #
# 4 # #
With 3 4s placed the final 4 is narrowed down to the top row and right column:
# # # 4
4 # # #
# # 4 #
# 4 # #
The initial insight showed the marked squares in the lower right quadrant are a 3 and a 4. With the 4 placed, the other marked square must be a 3:
# # # 4
4 # # #
# # 4 #
# 4 3 #
The marked squares in the lower left quadrant are 2, 3, 4. The 3 cannot be in the bottom row, leaving only 1 marked square for it, which in turn leaves only 1 marked square for the 2:
# # # 4
4 # # #
3 # 4 #
2 4 3 #
The only remaining square in the lower left quadrant must be a 1 (which was also clear from the initial insight that the other squares are 2, 3, 4). Similarly the only remaining square in the left hand column is a 1, and the only remaining squares in the last 2 rows must be a 2 and 1, in that order, which in turn means the remaining square in the right hand column is a 3:
1 # # 4
4 # # 3
3 1 4 2
2 4 3 1
The second row already has a 3 and a 4, and the second column already has a 1 and a 4, so the intersection of this row and column is a 2 (the only remaining number). This puts the 3 in the second column, and the 1 in the second row, narrowing down the last square to a 2:
1 3 2 4
4 2 1 3
3 1 4 2
2 4 3 1
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