Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
On my calculator I divided one whole number by another whole number and got the answer 3.125 If the numbers are both under 50, what are they?
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
We had many solutions to this challenge. This very clear explanation came from Roo at Okehampton Primary:
We found two solutions to the problem 'Fifteen Cards'.
$6, 9, 11, 12, 4, 14, 7$
$8, 7, 13, 10, 6, 12, 9$
We know we have found all of the ways because we went through them all by trying each number as our starting number. Here is our working:
$1, 14, 6, 17$ - no $17$ card
$2, 13, 7, 16$ - no $16$ card
$3, 12, 8, 15, 1, 17$ - no $17$ card
$4, 11, 9, 14, 2, 16$ - no $16$ card
$5, 10, 10$ - need to use $10$ twice
$6, 9, 11, 12, 4, 14, 7$ - IT WORKS!!
$7, 8, 12, 11, 5, 13, 8$ - need to use $8$ twice
$8, 7, 13, 10, 6, 12, 9$ - IT WORKS!!
$9, 6, 14, 9$ - need to use $9$ twice
$10, 5, 15, 8, 8$ - need to use $8$ twice
$11, 4, 16$ - no $16$ card
$12, 3, 17$ - no $17$ card
$13, 2, 18$ - no $18$ card
$14, 1, 19$ - no $19$ card
Well done - you worked in a very systematic way. Lyman from Lake Oswego Junior High School tackled the probelm in a similar way, but thought about the likely size of the first number.
The easiest way that unfortunately takes much time is to check the first card with all numbers from $1$ to $15$. Using advanced guessing skills, you could easily know that the first card is neither big, in this case $11-15$, nor small numbers, in this case $1-2$. The reason it's not a big number is because you will need a number bigger than $15$ in the third card because
placing a big number first makes you place a small number second and you don't have a card big enough to get the second and third cards' sum to be $20$.
It's neither small numbers because the second and third cards have a big sum and the third and fourth cards have a huge sum. Placing a small number first causes you to place a small number in the third card as a big number is in the second card, and you need a number bigger than $15$ in the fourth card.
The first card's solution to problems similar to this one tend to be around the mean of all numbers available. In this case it's $7.5$, so we should start checking from $7$ or $8$ and then $9$ and $6$, etc. Let's start with $7$. If the first card is $7$, the second had to be $8$ because $15-7=8$. The third card is $12$, as $20-8$ shows. The fourth card is $11$ since $23-12=11$. The next card
is $5$ since $16-11=5$. Using similar techniques, you easily get the sixth card is $13$ and the last card is $8$. However, it's not a solution because the last card needs to be same as the second card.
Another number with a good possibility for first card is $8$. So we'll check it. Using techniques same as when checking if $7$ is the first card, you should get the following results for the seven cards in order: $8, 7, 13, 10, 6, 12, 9$. Bingo. All numbers are different and less than $15$, and
consecutive numbers do have the sums given. We'll need to continue to check other numbers for the first card.
If you use $9$ as the first card, the seven cards in order are $9, 6, 14, 9$ etc. It's not a solution cause there are already two $9$s.
Maybe $6$ will work. The order of cards will be $6, 9, 11, 12, 4, 14, 7$. Another solution as all
are different, less than $15$, and have the consecutive sums.
Let's also check $5$. The order would be $5, 10, 10$ etc. It's obviously not the solution
as there are already two $10$s.
Now it's time to see $4$ as the first card. This time the order is $4, 11, 9, 14, 2, 16$ etc. It's also not the solution because there is a need for a card bigger than $15$.
Finally the only possibility left is $3$. The order of the seven cards should look like this: $3, 12, 8, 15, 1, 17$. It's also not the solution because it again involves a number bigger than $15$.
After eliminating the obvious incorrect choices, I have done minimal guess and check to find out there are two solutions to the problem. The first one I've found had the following seven cards in order: $8, 7, 13, 10, 6, 12, 9$. The other one I've found in the process had those seven cards in order: $6, 9, 11, 12, 4, 14, 7$. In both cases the biggest cards are third, fourth, and sixth
cards as I've thought.
Isabelle from South Wiltshire also used a very systematic approach but she started from somewhere different:
$23$ is the largest number that two cards have to add up to. $23$ can only be made by the pairs $15+8$, $14+9$, $13+10$ and $12+11$ so switching these around there are eight combinations so far. By trying out these eight combinations it is obvious that six of them do not work:
With the $15+8$ one there would have to be a second $8$ card to make the $16$.
With the $8+15$ one there would have to be a $1$ following the $15$ to make $16$ but then there would need to be a $17$ card to make the $18$ which is not possible.
With the $9+14$ one there would have to be a $2$ following the $14$ but then that would need a $16$ to make up the $18$ which is not possible as all cards must be $15$ or under.
The $14+9$ one requires a second $9$ as the first card.
The $12+11$ one does not work because it would need two $8$ cards and similarly the $10+13$ one would need a second $10$ card.
So the only remaining combinations are:
$8, 7, 13, 10, 6, 12, 9$
$6, 9, 11, 12, 4, 14, 7$
Caitlin from Hythehill Primary used a very similar approach to Isabelle. She said she drew a table with all the possible combinations for each total and she noticed that $23$ had the smallest number of possibilities.
Well done also to Owen from Montessori of Wooster Ohio, Year 7 at Loretto Junior School, Class 6 at St Martin's and Tom from Bridge and Patrixbourne who all sent clear solutions.