Formula Assistance: Take any two digit number (the two digits should be different). Write down all the different arrangements of the digits....

[mathteacher2019]

Hi Guys
Just signed up searching the forums to see if anybody else has come across this but havent had any luck so was wondering if anybody could assist.
Ive checked google searches, facebook, etc but no luck.
I have been working on a few different assignments / queries that have been posed but one has really stumped me.

So this is the query:

Take any two digit number (the two digits should be different).
Write down all the different arrangements of the digits. Add them up. Divide this total by the sum of the two digits.
What do you notice? Try again with a different two digit number. Why does this happen?
Extend this by looking at three / four digits/ Is there a way to predict the results?
Can you find a general formula? (What happens if you have repeated digits)?

So the bolded part is what is stumping me. From my working out (please do correct me if im wrong), I seem to find that for 2 digits, the total is always 11
(eg a = 7
b = 5
(5+7 = 12)
75 + 57 = 132 / 12 = 11)

For 3 digits, it seems that the total always equals 222.
From using a written down example, (eg 234), there are 6 different arrangements:
(234, 243, 342, 324, 432, 423)
4 + 2 + 3 = 9
Adding them all makes 1998 / 9 = 222

For 4 digits using same method, it shows 6666 as the total.

However the one part that is causing me an issue is trying to figure out a formula to use to figure out the answer when you choose the number of digits. Im unable to determine a pattern that will lead to finding this.

I tried for the simplest version (2 digits) but the best I could come up with is:

11(a+b) / b + a = 11
However that wouldnt really work in terms of an actual formula as it requires knowing the answer.
Is anybody able to assist please?

 

[Dr.Peterson]

To start, do you know a formula for the number of arrangements of an n-digit number with all digits different?

Then, think about what happens to the sum of digits when you add all those numbers.

 

[mathteacher2019]

To start, do you know a formula for the number of arrangements of an n-digit number with all digits different?

Then, think about what happens to the sum of digits when you add all those numbers.

Thanks for the reply.
So I tried that route and also struck out so wasnt sure it was the right path to finding the general formula for the initial part.

I had (for different digits used):
For 2 digits, there are 2 arrangements
For 3 digits, there are 6 arrangements
For 4 digits, there are 24 arrangements.
For n digits...

I couldnt think of a pattern here (was thinking to the power of, or it was linked to the previous digit, (n-1,etc)).
I was also thinking its based on the previous digits arrangements.
So for example 3 digits - you would use number of arrangements from previous number (2)
So 3 x 2 = 6.
And for 4 digits, previous numbers (3) arrangement is 6, so 4 x 6 = 24
Thus thinking for 5 digits, it would be 5 x 24 = 120
But this is reliant on knowing the previous digits number of arrangements.

Am I missing something completely obvious as I feel like im just staring at it.

I do see a sort of pattern
eg for 2 digits, there are 2 arrangements.
And total is 11

Using number of arrangements from previous number (2), if I use number of digits (3), total would be 2 three times in a row (222)
For 4, it would be previous number arrangement (6) typed out four times in a row (4444)

However I cant seem to put it together. I feel like i overcomplicating it or just missing something completely.

Sorry if all seems a mess in terms of explanation. I cant seem to simplify this particular conundrum.

 

[ksdhart2]

The number of ways to rearrange an \(n\)-digit number where order matters and repetition is not allowed is given by \(n!\), read aloud as n factorial. Intuitively this makes sense if we think about how many choices we have for each digit. The first digit can be any of the \(n\) choices. That leaves \(n-1\) choices for the second digit, \(n-2\) choices for the third digit, ..., all the way down to two choices for the second-to-last digit and only one choice for the last digit. That gives us a total of \(n(n - 1)(n - 2) \cdots (3)(2)(1) = n!\) ways to arrange the digits.

Now let's look at the case for a two-digit number. Would you agree that a generic two digit number is \(10a + b\), where \(a\) and \(b\) are the digits of the number? The other arrangement of these digits would then be \(10b + a\), so their sum is \(10a + b + 10b + a = 11a + 11b\). And the sum of the digits is just \(a + b\). That means the ratio of these two things must be \(\frac{11a + 11b}{a + b} = \frac{11(a + b)}{a + b} = 11\).

The case for a three-digit number is a bit more complicated, but the same basic principle applies. A generic three digit number is \(100a + 10b + c\), again where \(a, \: b, \: c\) are the digits of the number. Writing out all six arrangements:

• 100a + 10b + c
• 100a + b + 10c
• 10a + 100b + c
• 10a + b + 100c
• a + 100b + 10c
• a + 10b + 100c

We see that their sum must be \(222(a + b + c)\), and their digits must sum to \(a + b + c\) which yields an answer of 222. From this, can you see why your result that a four-digit number always yields a result of 6666 makes perfect sense, and fits the pattern? What do you suppose a five-digit number would yield? An \(n\)-digit number?

 

[mathteacher2019]

Hi @ksdhart2
Thanks for explaining. I tried to work through this the last few days but I just cant crack it.

I get what youre saying on the arrangements but for me it seems, like you can only get the the formula from writing it all down as arrangements possible. eg
for the 3 digit, if I write it out, sure, I can get it and see that its 222 (a + b + c)
And likewise for for 4 digit, its 6666 (a + b + c + d)
But thats through working it out.

I cant see see how to do this if I pick a big random number eg like 9 without knowing the previous numbers or the specific arrangements.

For example, how do I figure out the 6666 without writing down the options?
The n! (factorial) makes sense in terms of figuring out the arrangements.
Im struggling with putting that into a n type formula for the value (when arrangements all have to be different though)

Maybe it keeps going over my head and its just not clicking (or ill never get??)
I did spend a few days trying to work through this and also asked a colleague but I think they didnt get it hence complicating it further for me.

 

[Steven G]

Hi Guys
Just signed up searching the forums to see if anybody else has come across this but havent had any luck so was wondering if anybody could assist.
Ive checked google searches, facebook, etc but no luck.
I have been working on a few different assignments / queries that have been posed but one has really stumped me.

So this is the query:

Take any two digit number (the two digits should be different).
Write down all the different arrangements of the digits. Add them up. Divide this total by the sum of the two digits.
What do you notice? Try again with a different two digit number. Why does this happen?
Extend this by looking at three / four digits/ Is there a way to predict the results?
Can you find a general formula? (What happens if you have repeated digits)?

So the bolded part is what is stumping me. From my working out (please do correct me if im wrong), I seem to find that for 2 digits, the total is always 11
(eg a = 7
b = 5
(5+7 = 12)
75 + 57 = 132 / 12 = 11)

For 3 digits, it seems that the total always equals 222.
From using a written down example, (eg 234), there are 6 different arrangements:
(234, 243, 342, 324, 432, 423)
4 + 2 + 3 = 9
Adding them all makes 1998 / 9 = 222

For 4 digits using same method, it shows 6666 as the total.

However the one part that is causing me an issue is trying to figure out a formula to use to figure out the answer when you choose the number of digits. Im unable to determine a pattern that will lead to finding this.

I tried for the simplest version (2 digits) but the best I could come up with is:

11(a+b) / b + a = 11
However that wouldnt really work in terms of an actual formula as it requires knowing the answer.
Is anybody able to assist please?

Just some comments on what you wrote.
75 + 57 = 132 / 12 = 11 No, 75 + 57 does not equal 132/12.
11(a+b) / b + a = 11 No, not true. It is true that 11(a+b) / (b + a)= 11

 

[Steven G]

Thanks for the reply.
So I tried that route and also struck out so wasnt sure it was the right path to finding the general formula for the initial part.

I had (for different digits used):
For 2 digits, there are 2 arrangements 2!
For 3 digits, there are 6 arrangements 2*3=3!
For 4 digits, there are 24 arrangements. 3!*4 =4!
For n digits... =n!

See red above

 

[Steven G]

Let abc represent 100a + 10b + c
The six permutations are:
abc
acb
bac
bca
cab
cba

Note that each column adds up to 2(a+b+c)

So the sum of the 6 numbers is 100*2(a+b+c) + 10*2(a+b+c) + 1*2(a+b+c)= 111*2(a+b+c) = 222(a+b+c).

It wasn't that hard!

Now find out why 6666!

 

[Steven G]

Hi @ksdhart2
Thanks for explaining. I tried to work through this the last few days but I just cant crack it.

I get what youre saying on the arrangements but for me it seems, like you can only get the the formula from writing it all down as arrangements possible. eg
for the 3 digit, if I write it out, sure, I can get it and see that its 222 (a + b + c)
And likewise for for 4 digit, its 6666 (a + b + c + d)
But thats through working it out.

I cant see see how to do this if I pick a big random number eg like 9 without knowing the previous numbers or the specific arrangements.

For example, how do I figure out the 6666 without writing down the options?
The n! (factorial) makes sense in terms of figuring out the arrangements.
Im struggling with putting that into a n type formula for the value (when arrangements all have to be different though)

Maybe it keeps going over my head and its just not clicking (or ill never get??)
I did spend a few days trying to work through this and also asked a colleague but I think they didnt get it hence complicating it further for me.

OK, I see now that you know how to get 6666 but want to get it more easily. So you need to multiply 1111 by 6. I assume that you know how to get the 1111. So how do you get 6? Well you will have 4! different numbers that you will be adding. But each column will have 4 different digits. So how many times will each digit come up? Answer will be 4!/4 = 3! =6 (!!!!!)