We've created Halloween-themed math puzzles that are perfect for any age group. All your child needs in order to solve them is their curiosity, logical thinking, and the ability to multiply and divide numbers! Keep the joy of discovery alive by providing hints (which we have included below) instead of explaining the answer.
As you pass by a neighbor’s house, you notice that there is one jack-o-lantern on the porch and another on the railing. Both jack-o-lanterns are lit, but what if we wanted to vary which ones are lit every day based on our mood? How many days will it take until no new combination can be made?
It becomes clear quickly that with 2 pumpkins, there are only 4 different combinations:
1. Both pumpkins are lit;
2. Only the pumpkin on the porch floor is lit;
3. Only the pumpkin on the railing is lit;
4. Neither pumpkins are lit;
What else can we do? What if we add more pumpkins? Let’s add three more for now! We put 1 pumpkin on the porch, 1 on the first step of the stairs, and we put another 2 pumpkins on the second and third steps of the stairs. Will there be enough combinations for a month?
How does the number of possible combinations change by adding just 1 pumpkin?
After adding 1 pumpkin, the number of lighting combinations possible doubles: All the previous combinations stay, and the new pumpkin is not lit. When the new pumpkin is lit, from every old combination you can get a whole new one by including the new pumpkin that is “on.”
8 combinations for 3 pumpkins, 16 combinations for 4 pumpkins and 32 combinations for 5 pumpkins.
At this moment, a mathematician feels the overwhelming desire to generalize this solution to any number of pumpkins! Isn’t it obvious that for 6, 7, 100 and 1,000 pumpkins the algorithm will be the same? For every new pumpkin, the number of possible combinations will double. So the answer is 2 times 2 for as many times as there are pumpkins! For instance, in the case of 5 pumpkins, the answer is 2 x 2 x 2 x 2 x 2. For 6 pumpkins, the answer is 2 x 2 x 2 x 2 x 2 x 2, and so on.
Great! With just 5 pumpkins we will have 32 combinations, that's enough for over a month! Let's add in a twist: what if we added in some artistic preferences?
If right now there are a few (not necessarily an odd number) of pumpkins lit, then we can turn them all off and then light up the rest of the pumpkins. This will give us a new combination.
Let's divide every combination into pairs: If for one combination there are a few pumpkins lit, then its pair will have the rest of the pumpkins lit. From the last puzzle, we already know the amount of combinations for 5 pumpkins, 32. Because of this, we know that there will be 16 pairs. In every pair, there will be 1 with an even number of lit up pumpkins, and 1 with an odd number of lit up pumpkins. So this would give us a total of 16 combinations.
16 combinations
It's easy to find every possible combination of lit up pumpkins (like we did in Puzzle 1): If we add 1 more pumpkin, then the previous number of combinations would be doubled and we would get 32x2=64. Again, we can do what we did in the previous puzzle and divide every combination into complementary pairs, but here is the problem: this time either inside each pair of combinations there will be an even number of pumpkins lit up for both, or there will be an odd number. The last way to find the solution does not work, we need to find a new one!
Let's light up the first 5 pumpkins whichever way we want (in any combination, not watching whether it is odd or even). We can make the 6th pumpkin “controlled:” turning it on last and only if in the first 5, there is an even number of pumpkins lit up.
Let's break up the combinations of lighting into 32 pairs. We have 32 combinations of lighting for the first 5 pumpkins. From each combination, we get two possibilities (it can either be “on,” or not) for 6 pumpkins. With such a division into pairs in exactly one of the two variants of the pair, an odd number of pumpkins will be “on.” Therefore, there are as many combinations as there are pairs.
32 ways
We can make this exact pumpkin (on the third step) the “controlled pumpkin”
We will make the pumpkin on the third step the “controlled” variable: so we will light it last and only if – not counting this pumpkin – there is an even amount of lit pumpkins. It turns out that this pumpkin will be “on” for exactly as many days as there are ways to include an even number of pumpkins from the remaining 5. We will calculate the number of these combinations: 32-16=16. (32 is the number of lighting options with 5 pumpkins, and 16 is the number of options to include an odd number of 5 pumpkins, we considered this in puzzles 1 and 2. We subtract the second answer from the first.)
16 days
If you got to this puzzle and figured out the previous ones, then well done!
Once you start on your math journey you will find yourself asking more and more questions, and trying to figure out the answers! Your curiosity will continue to drive you forward to learn more and more.
We have one final problem for you as a prize for getting this far. Have fun solving it! Since you made it this far, you don't need the solution, just a hint!
Make not one, but several pumpkins “controlled.”
