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Stage: 1 and 2 Challenge Level: Challenge Level:1

We had quite a few responses from pupils who had found out things by exploring this situation. Calum, Christopher and Matthew from St. Andrews in Scotland wrote to say:

 

We found out that using the digits $1, 2$ and $3$ making a four-digit number you would start doubling eventually when you reach line $10$ every time no matter what order you put the digits in.
We found out that using the digits $1, 2$ and $4$ making a four-digit number you would start doubling eventually when you reach line $7$ every time no matter what order you put the digits in.
We found out that using the digits $1, 3$ and $4$ making a four-digit number you would also start doubling eventually when you reach line $7$ every time no matter what order you put the digits in.
We found out that using the digits $2, 3$ and $4$ making a four-digit number you would start doubling eventually when you reach line $6$ every time no matter what order you put the digits in.  

 

Sayeed from St. Michael's London also sent in a well thought out reponse:-

This is my answer when starting with $4$ digits:

$3243,  122314,  21221314,  31321314,  31123314,  31123314$

It will continue as the same number forever.

I notice: Each row has two more digits then the previous row until the rows have the maximum amount of digits ($8$) possible. In each row with the maximum digits, the second digit always has a $1$, the fourth digit always has a $2$, and the sixth digit always has a $3$ and so on. Every row ends in $14$ excluding the starting row. This is my answer when starting with $5$ digits:

$22411,  212214,  213214,  21221314,  31321314,  31123314,  31123314$

It will continue as the same number forever.

I notice the same thing that happens with the starting $4$ digit row except the rows getting two more digits each time until there are maximum digits. I also notice that with the starting four digit row it takes four counts till the number continues as the same number forever. But for the starting five digit row it takes five counts till the number continues as the same number forever!  

Thomas from Colet Court School said the following and attached his numbers.

When you start with $4$ digits the series converges to $21322314$ whatever the first $4$ digits.  When you start with $5$ digits it gives the same result.  
 
 Thomas
 

Finally, Miss Stanley's Numeracy group from Greystoke Leicester wrote:-

 

We liked this challenge and worked very hard. Trying other numbers using the same rules we found that we could continue until the numbers were the same, because that number would keep repeating. We discovered some have shorter sequences and some have longer sequences until the same number repeats, but we're not sure why yet.
Caitlin and Millie found that some numbers ($4122$) didn't seem to have an end because we spotted the pattern that it kept repeating itself, so we decided to stop. Some of us even moved onto extending this challenge to $5, 6$,and even $7$ digits.
Many of us spotted that the larger the number of digits in the starting number, the shorter the sequence was to get to the end.
Thank you, we enjoyed this challenge.

 

 Well done to all the contributors, it sounds as if you really enjoyed this. (A late arrival came from Adam at Cypress School who noted something special about $1$ & $4$.  I am wondering if that was because it was not a usual thing you'd find going on in many Mathematics lessons? On the last day of the month we recieved this excellent presentation of Oscar from Spain.

You have to take out one digit of the $1,2,3,4$ which are the possible digits to make the starting number. If you take out a number and want to get a $4$ digit number, you have to repeat one of the other $3$ numbers. If you take out $1$, you have possible starting numbers
$2234, 2334, 2344$ and other possible numbers that you get changing the order of the digits in each of those $3$ numbers.
As the order does not affect digit counting, those give the same counting sequence. The counting is:
$A    2    B    3    C    4$
and A-B-C have to be $2-1-1$ (for $2234$) or $1-2-1$ (for $2334$) or $1-1-2$ (for $2344$).
The next counting in all cases is $2 1 2 2 1 3 1 4$ and sequence is:
$3 1 3 2 1 3 1 4$
$3 1 1 2 3 3 1 4$ and this last number stays the same if you count the digits. If you take out $2$ you get the number:$ 2 1 3 2 2 3 1 4$ If you take out $3$ you get the same as if you take out $1$, and if you take out $4$ you get the same as if you take out $2$.