Fifteen cards

7506 /www/nrich/html/content/id/7506/ 1 2011-07-26T09:26:06 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> I have fifteen cards numbered $1 - 15$. I put down seven of them on the table in a row. <mdo:image alt="7 cards" height="92" src="Cards.png" width="511"></mdo:image> The numbers on the first two cards add to $15$. The numbers on the second and third cards add to $20$. The numbers on the third and fourth cards add to $23$. The numbers on the fourth and fifth cards add to $16$. The numbers on the fifth and sixth cards add to $18$. The numbers on the sixth and seventh cards add to $21$. What are my cards? Can you find any other solutions? How do you know you've found all the different solutions?</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Fifteen Cards</h2> I have fifteen cards numbered $1 - 15$. I put down seven of them on the table in a row. <mdo:image alt="7 cards" height="92" src="Cards.png" width="511"></mdo:image> The numbers on the first two cards add to $15$. The numbers on the second and third cards add to $20$. The numbers on the third and fourth cards add to $23$. The numbers on the fourth and fifth cards add to $16$. The numbers on the fifth and sixth cards add to $18$. The numbers on the sixth and seventh cards add to $21$. What are my cards? Can you find any other solutions? How do you know you've found all the different solutions? </div> <h3>Why do this problem?</h3> <a href="http://nrich.maths.org/7506&part=">This problem</a> requires no mathematical ideas beyond simple addition and possibly subtraction, but it does require the perseverence to stick with a trial and improvement approach, combined with some systematic working. <h3>Possible approach</h3> <div>You could start by giving pairs of learners a set of cards numbered from $1$ to $15$. (Numbered cards can be downloaded <a href="/content/id/7506/15Cards.pdf">here</a>. If they are printed onto thin card they will be easier to use and if covered with plastic film they should last a long time.) Suggest they put seven cards face down in a row at random and find out the sum of the numbers on the first two cards. You could find out whether any pairs had the same total and ask the rest of the class to predict whether they had used the same numbers. Invite children to justify their responses, for example by listing possible combinations to make that sum. </div> <div> </div> <div>Having enabled learners to get a feel for the situation in this way, introduce the problem itself, perhaps by giving out copies of <a href="/content/id/7506/15Cards2.doc">this printed sheet</a> or by displaying the challenge on the board. Try not to say anything else at this stage and give them chance to work in their pairs so that they are able to talk through their ideas with their partner.</div> <div> </div> <div>Observe the way that pairs work and, once they have made some headway, bring the group back together to discuss strategies so far. What have they been doing to try and solve the problem? Some may have adopted a trial and improvement approach. Others may be working more systematically, perhaps by listing possible pairs of numbers for a particular total. Can anyone suggest a good place to start the problem? Has anyone tried something that didn't work? After this discussion, give more time for pairs to complete the task. You may find that some adopt a strategy suggested by another pair rather than pursuing their original method.</div> <div> </div> <div>In the final plenary, come together again to reflect on the process. You could share final answers, but also encourage some pairs to explain their methods from start to finish. This could be an opportunity to discuss which method/s were particularly elegant or efficient.</div> <div> </div> <h3>Key questions</h3> <div>How might you start this problem? </div> <div>How could this number be made with two cards? Is there another way? And another ...?</div> <div>How will you remember which combinations you have tried?</div> <div>I wonder whether the order of these two cards might make a difference?</div> <div> </div> <h3>Possible extension</h3> After demonstrating that they have found all the possible solutions, learners could make up a similar problem for others to try. Remind them that you will expect them to know the solutions to their own problem before giving it to others to try out! <h3>Possible support</h3> Using digit cards will encourage learners to try out different combinations without having to commit anything to paper at first. They may need reminding that, for example, that $12$ followed by $3$ will give a different order from $3$ followed by $12$. You could suggest that they focus on just one pair to begin with and consider all possible combinations, then try to work out what the other cards could be based on each of those possibilities. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Remember that there is only one of each number from $1 - 15$. You can download a set of cards numbered $1 - 15$ <a href="/content/id/7506/15Cards.pdf">here</a>. How will you remember which combinations you have tried? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Two solutions: 8, 7, 13, 10, 6, 12, 9 6, 9, 11, 12, 4, 14, 7 </mdoxml> 5 3 0 1 0 0 0 Fifteen cards I have fifteen cards numbered $1 - 15$. I put down seven of them face down in a row. Can you use the information to find out which cards they are? Using, Applying and Reasoning about Mathematics Working systematically Using, Applying and Reasoning about Mathematics Selecting and using information Using, Applying and Reasoning about Mathematics Trial and improvement Calculations and Numerical Methods Addition & subtraction Mathematics Tools Digit cards Admin Upper primary mapping document