Dad needs help.

[Chrispick77]

Please help me, my 9 yr old daughter came home with a challenge from school (it's not homework).
I want to explain it to her but I need to know I'm doing it correctly.
The challenge was

Using the digits 1-8 how many different ways can the digits be arranged to make 2 numbers whose sum is 9999.

So i take it they want two numbers like 1234 and 8765.
I originally thought it was 24 but was only rearranging 1234 and 5678 then realised I could swap say the 8 and 4 and make more.
I've now got to 54 and I still think I'm nowhere near.
Is there a quicker way to work this out as I now want the answer myself and want to explain how to do it so she can actually do it herself.

Chris

 

[Dr.Peterson]

Assuming you can convince yourself that there can be no carries in the addition, it amounts to assigning one of the pairs 1+8, 2+7, 3+6, 4+5 to each column, and choosing which goes on top in each column.

 

[JeffM]

My answer is very similar to Dr. Peterson's.

First, the largest number under 9999 that you can make using all 8's is 8888 so the smallest that will suit is 1111. Your conclusion that both numbers must contain four digits is correct.

Second, we need to consider whether we need to worry about "carrying." There can't be any carrying from the units digits because the sum of two positive numbers smaller than 8 is either under 10 or under 17. So the only way we can get a 9 in the units digit of the sum is for the two units digits to sum to exactly 9. Got that? Does that logic extend to the tens and hundreds digits?

Third, how many ways can you create a four digit number > 1110 and < 8889? Let's see? You have 8 ways to choose the thousands digit. For each of those, how many ways can you choose the hundreds digit?

Fourth, given any suitable first number, how many ways can you choose the second?

So the total number of pairs is what?

This is really a problem in understanding place value and systematic counting.

 

[pka]

Frankly, I find the question vague. Can the pair \(\left\{ \begin{gathered} 8271 \hfill \\ 1728 \hfill \\ \end{gathered} \right\}\) be used? I suspect not.
Putting on a editor's cap, I think it means: Using each of the digits 1 to 8 once \(\cdots\).

 

[Chrispick77]

Frankly, I find the question vague. Can the pair \(\left\{ \begin{gathered} 8271 \hfill \\ 1728 \hfill \\ \end{gathered} \right\}\) be used? I suspect not.
Putting on a editor's cap, I think it means: Using each of the digits 1 to 8 once \(\cdots\).

This is how I read it. Each digit once.

 

[Dr.Peterson]

This is how I read it. Each digit once.

Have you (or, better, your daughter) tried the methods from posts #2 or #3? Once you do that, we'll be able to check your thoughts, and probably take it to a conclusion.