10 square matchstick puzzle
Remove 5 matches to leave 5 squares in the puzzle figure comprising as many as 10 squares of same size. No match must be left hanging.
The Puzzle
The following matchstick figure comprises many squares. Remove 5 matches to leave just 5 squares.
Recommended time to solve is 25 minutes.
Though not easy, try to solve this interesting puzzle. It will sure be fun.
Solution to remove 5 matches to leave 5 squares puzzle: Count of sticks and squares
First, you count the number of matchsticks and the exact number of squares. Number of sticks is 27 that make the squares.
While counting the squares, the squares of different sizes also counted. Total number of small squares is 10 whereas the number of large squares (of 2-stick side length) is four. The following figure shows the formation of the large squares. The three large squares are identified by coloring the outlines in brick red, whereas the fourth large square comprises the small squares A, B, C and D.
Apart from these four large squares, the figure has 10 more small squares—a total of 14 squares.
Because of this large number of squares, along with many common sticks, it is essential to have a clear idea of all the squares in the figure at the start itself.
Solution to remove 5 matches to leave 5 squares puzzle: Trying to get an idea of the solution figure using a trial
A few quick intuitive attempts failed to produce the desired solution. The puzzle is not easy. It needs an analytical approach. Random attempts will not be of any use.
First question we ask—can we form any general rule on the solution process? The number of squares is large. It is 14. By removing 5 matches to you have to reduce the number to only 5.
Nine squares are to be eliminated by removing 5 matches.
It means,
Conclusion 1: Each stick removal must be highly efficient in eliminating squares of two different sizes.
This concept of square removal efficiency should play an important role in solving the puzzle analytically.
In 5 moves, you have to eliminate 9 squares. On an average, you have to eliminate about 2 squares with each stick removal.
You may conclude with reasonable certainty,
Conclusion 2: For the solution, you cannot remove any corner stick. You eliminate only one square by removing a pair of corner sticks. Square removal efficiency is too low for a pair of corner sticks (you cannot remove one corner stick only because it will create one hanging stick).
It follows,
Conclusion 3: For the solution, only common sticks should be removed, and that too only common sticks that eliminate the maximum of squares.
These conclusions follow commonsense reasoning as well.
With more understanding about what type of sticks to remove, we get down to form our first trial. The following shows the trial figure of 5 squares.
Six matches removed for creating 5 squares. We could not get to the solution in this trial, but leaned a good deal about it.
First action we took is to remove the sticks 1 and 2. We have identified this pair as a highly favorable pair of sticks because removing these two eliminated 3 small squares and 2 large squares. Removing sticks 3 and 4 next also eliminated 3 small squares. Together, removing these 4 sticks eliminated 8 squares—a very good achievement no doubt.
Because of this highly effective result, we identify each of the two similar structures of one small square attached inside in one corner of a large square as very favorable that should be part of the solution.
But alas! Even with this great effort, we didn't get the solution. To eliminate the ninth square, we had to remove any of the pairs of corner sticks 5, 6 or 7, 8. The total number of sticks removed becomes 6 instead of 5.
Though we failed to create the solution, we got a good feel about it by the trial.
Solution to the remove 5 sticks to leave 5 squares: Further structural analysis and reasoning
First question we asked: why did we fail? The reason identified is clear:
Reason of failure in the first trial: At the last step, a pair of corner sticks were removed that was prohibited.
It follows,
Conclusion 4: For the solution, besides the highly efficient pair of configuration of one small square inside a larger one, you must remove the single common stick at the top.
Let us show the result of the first two steps of the trial for better understanding.
In the solution, we must remove the common stick 9 instead of removing two of the corner sticks, 5, 6 or 7, 8.
But we cannot do it with our first two-step result of the trial, though the configuration achieved has been highly efficient.
At this point, it takes just one more step to reorient the squares A and B to move them up and position the two as adjacent squares inside the two large squares below the top two squares. This is the perfect situation to remove the common stick 9 and get the solution as shown.
The fifth square is no longer a small square. It is the third large square at the top earlier shown. It intersects with the two adjacent large squares below to create two additional small squares. Total number of small squares becomes thus 5. Matchsticks removed are sticks 1, 2, 3, 4 and 5.
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