Curious number

7218 /www/nrich/html/content/id/7218/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <div style="text-align: center;"><mdo:image width="287" height="118" src="7218.png" alt="1-6"></mdo:image></div> Can you order the digits $1, 2, 3, 4, 5$ and $6$ to make a number which is divisible by $6$ ... ... so that when the final or last digit is removed it becomes a $5$-figure number divisible by $5$? And when the final digit is removed again it becomes a $4$-figure number divisible by $4$? And when the final digit is removed again it becomes a $3$-figure number divisible by $3$? And when the final digit is removed again it becomes a $2$-figure number divisible by $2$, then finally a $1$-figure number divisible by $1$? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Thank you for your solutions to this tricky challenge. Sebastian and Josh from Swaffham Prior C of E Primary School Mega Maths Club found two solutions to the six-digit challenge and outlined their method for us: <div>$123654$ (Sebastian) and $321654$ (Josh)</div> <div> </div> We knew that the fifth number had to be the $5$. We also knew that the second, fourth and sixth numbers had to be even because they had to be multiples of even numbers. This meant that the first and third numbers could only be either $1$ or $3$. Knowing this we used trial and error to find a number that followed the rules. We used a calculator to find out if the numbers were multiples of $3$, $4$ and $6$. We didn't need to use a calculator for the first, second and fifth numbers as these were obvious. Zara from Oldfield Primary School gave us a bit more detail about how she worked on the solution. First of all we decided to give each digit a position number eg: for the number $235614$ Position $1$ = $2$, Position $3$ = $5$ Next we worked out that position $5$ must be a $5$, because it has to be divisible by $5$ and there is no zero. We then saw that position $2$, $4$ and $6$ must be even numbers, because they had to be divided by $2$, $4$ or $6$. That left position $1$ and $3$, which had to be $1$ or $3$! We knew that a number is divisible by $3$ if its digits when added together are divisible by $3$. So, the first three numbers must be $1$, $3$ and then an even number in position $2$, so when they are added together are divisible by $3$. We tried $1 + 4 + 3 = 8$ this is not divisible by $3$ We tried $1 + 6 + 3 = 10$ this is not divisible by $3$ We tried $1 + 2 + 3 = 6$ this IS divisible by $3$ These are the first three numbers - and we found out that $1$ and $3$ can be swapped around $123$ or $321$. We now had $123?5?$ We worked out the last two numbers by trying them and ended up with solutions: $123654$ and $321654$. Daniel also explained very clearly how he reached the solution. He started in the same way as Zara but then used a slightly different approach: Since for the first operation (i.e. dividing the six-digit number by $6$) the number must be divisible by $6$, the last single digit number must be even. Since for the second operation (i.e. dividing the five-digit number by $5$) the number must be divisible by $5$, the second last single digit number can only be $5$. Since for the third operation (i.e. dividing the four-digit number by $4$) the number must be divisible by $4$, the third last single digit number must be even. Since for the fifth operation (i.e. dividing the two-digit number by $2$) the number must be divisible by $2$, the fifth last single digit number must be even. By process of elimination, the other two numbers must be $1$ and $3$. So for the format of the six-digit number, we have: ($1$ or $3$),($2$, $4$ or $6$),($1$ or $3$),($2$, $4$ or $6$),$5$,($2$, $4$ or $6$) For the three-digit number to be divisible by $3$, its digits must add up to a multiple of $3$. Since this number contains $1$ and $3$, along with $2$, $4$ OR$ $6 we can work out: $1+3+2=6$ (divisible by $3$) $1+3+4=8$ (not divisible by $3$) $1+3+6=10$ (not divisible by $3$) Therefore the second number must be $2$. So the format of the number is now: ($1$ or $3$),$2$,($1$ or $3$),($4$ or $6$),$5$,($4$ or $6$) Since for the third operation (i.e. dividing the four-digit number by $4$) the number must be divisible by $4$, the number made by its last two digits must also be divisible by $4$. These digits are ($1$ or $3$) and ($4$ or $6$) Therefore the possible numbers are $14$, $16$, $34$, and $36$ Since only $16$ and $36$ are divisible by $4$, the fourth digit of the number must be $6$. By process of elimination the last number must therefore be $4$. So we are left with ($1$ or $3$),$2$,($1$ or $3$),$6$,$5$,$4$ This gives the two numbers $123654$ and $321654$, both of which are correct, as shown above. Hannah, Megan, Jess, Toby, Callum, Callum, Lydia, Nadia, Sam and Chris from Peterchurch Primary School also used very similar reasoning to start with, but used a different method again towards the end of their solution: $123654$ and $321654$ We worked this out by the following process: 1. We knew that the fifth digit must be a $5$ because that's the only number divisible by $5$ (because we had no zero) 2. We knew that the second, fourth and sixth digits had to be even to be divisible by $2$, $4$ and $6$. 3. The first and third digits had to odd - $1$ or $3$ 4. We worked out that the second digit must be $2$ because 143 or 163 would not be divisible by $3$ 5. This meant that we only had four combinations left: $123456$, $123654$, $321456$, $321654$ We tested these to find the only two possible solutions. We also received fantastic solutions from Samuel at St Michael's Primary in Oxford; Orange Kingfishers at Wood Farm Primary; Alyssa, Caleb, Chelsea, Jake and Katie from Myland Primary and Krystof from Uhelny Trh, Prague. Finally, Rachel from Greenacre School for Girls took the challenge even further. She used digits $0$ to $9$ and found one solution: $3,816,547,290$ I used trial and improvement to solve this - it took me about two hours! Very well done, Rachel. What perseverence! The challenge that Rachel set herself is in fact another NRICH problem in its own right: <a href="http://nrich.maths.org/796&part=" class="editorial">American Billions</a>. Try it yourself! </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Curious Number</h2> <div style="text-align: center;"><mdo:image alt="1-6" height="118" src="7218.png" width="287"></mdo:image></div> Can you order the digits $1, 2, 3, 4, 5$ and $6$ to make a number which is divisible by $6$ ... ... so that when the final or last digit is removed it becomes a $5$-figure number divisible by $5$? And when the final digit is removed again it becomes a $4$-figure number divisible by $4$? And when the final digit is removed again it becomes a $3$-figure number divisible by $3$? And when the final digit is removed again it becomes a $2$-figure number divisible by $2$, then finally a $1$-figure number divisible by $1$? </div> <h3>Why do this problem?</h3> <a href="http://nrich.maths.org/7218&part=">This problem</a> is intentionally difficult for Stage 2 as it stands. However, these notes are written so that simplified versions are tried first which means insights can then be applied to the problem itself. This therefore presents opportunities for discussing efficient or elegant solutions and working in a systematic way. Some knowledge of divisibility rules is essential. <h3>Possible approach</h3> <div>Write up the number $123$ (one hundred and twenty three) on the board for all to see and give time for learners to talk together about anything they notice or know about the number. Share some suggestions, which might include, for example, that it is odd, it has three digits, the digits are consecutive, it is a multiple of three ... There will be many ideas hopefully!</div> <div> </div> <div>Explain that this is in fact a special number. If you look at the number created by all three digits ($123$), it is divisible by $3$. If you look at the number created by just the first two digits ($12$), it is divisible by $2$. If you look at just the first digit ($1$), it is divisible by $1$. Say something along the lines of "I wonder whether there are any other ways of arranging the digits $1$, $2$ and $3$ to create a number which has these same properties?" In other words, the challenge is to order the digits $1$, $2$ and $3$ to make a number which is divisible by $3$ ... so that when the final digit is removed it becomes a two-figure number divisible by $2$ ... and when the final digit is removed again it becomes a one-figure number divisible by $1$.</div> <div> </div> <div>It might be worth beginning this briefly on the board so that everyone is very clear what the question means: You could suggest that another way of ordering the numbers $1$, $2$ and $3$ is, for example, $2$ then $3$ then $1$ to make the number two hundred and thirty one ($231$). So, is the number $231$ divisible by $3$? How do they know? [Yes, it is. Some children might mentally perform a division calculation. Others might know that if the digits add to a number that is divisible by $3$, then the whole number is divisible by $3$.] So far, so good. Taking off the last digit, we're left with the number $23$. Is this divisible by $2$? [No - it's odd.] So, the number $231$ doesn't meet all the requirements. At this point, invite pairs or small groups to continue working on this problem and give them a taste of what their next challenges will be: Can they do the same for four-digit numbers using the digits $1$, $2$, $3$ and $4$? How about five-digit numbers using $1$, $2$, $3$, $4$ and $5$? A six-digit number ... etc?</div> <div> </div> <div>Once the children have had a go at the three-digit and perhaps four-digit possibilities, bring them together to talk about their methods. This will bring up the fact that some digits have to go in certain places. For example, which places have to have even numbers? This might also be a good opportunity to discuss other divisibility rules. Is there a method which stands out as being particular efficient? Or is there a method that some children particularly like for whatever reason? (NB It is not possible to find a four-digit or a five-digit number that follow these rules but you can expect the children to be able to convince you this is the case.) After this discussion, learners could continue to work on the problem and they may well change the method they use. </div> <div> </div> <div>The plenary could involve sharing solutions to the six-digit problem but, in particular, invite the children to comment on how having a go at simpler versions of the task helped them tackle the six-digit (or more) challenge. How do they think they might have got on if they'd have been presented with the six-digit version at the start? What does this tell us about solving mathematical problems?</div> <h3>Key questions</h3> <div>What makes a number divisible by one/two/three/four/five/six ...?</div> <div>Where do the even numbers have to go?</div> <div>So where do the odd numbers have to go?</div> <div>Where does the five have to go? </div> <h3>Possible extension</h3> Learners could attempt to order the ten digits from $0$ - $9$ in the same way which is the basis of the very challenging <a href="http://nrich.maths.org/796&part=">American Billions</a> problem. <h3>Possible support</h3> The tables on <a href="/content/id/7218/7218.pdf">this sheet</a> might help learners organise their working - the columns can represent place value. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Do you know any divisibility rules? A number is divisible by $3$ if its digits, when added together, are divisible by $3$. For example, take $174: 1 + 7 + 4 = 12$ which is divisible by $3$. You can add it as many times as you want. $12: 1 + 2 = 3$ A number is divisible by $6$ if it is an even number and it is divisible by $3$. A number is divisible by $4$, if the tens and units form a number which is divisible by $4$, for example $732$ and $9048$ are divisible by $4$ (because $32$ and $48$ are divisible by $4$, but $338$ and $2342$ are not (because $38$ and $42$ are not divisible by $4$). (Why does this work?) It could be a good idea to make a table to keep track of where the digits $1$ to $6$ could go. Where will the even numbers have to go? So what about the odd numbers? Where will the $5$ have to go? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">123 and 321for three-digit impossible for a four-digit number impossible for a five-digit number 123654 and 321654 for six-digit </mdoxml> 5 5 0 1 0 0 0 Curious number Can you order the digits from 1-6 to make a number which is divisible by 6 so when the last digit is removed it becomes a 5-figure number divisible by 5, and so on? Using, Applying and Reasoning about Mathematics Working systematically Numbers and the Number System Factors and multiples Numbers and the Number System Odd and even numbers Numbers and the Number System Divisibility