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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.