Here’s a puzzle, from (believe it or not) WhiteHouseKids.gov

One Through Five

• Consider all five-digit numbers formed by using the digits 1, 2, 3, 4, and 5 once and only once in each number formed. What is the sum of all of these five-digit numbers?

You can punch your answer in at the site above to see if you got it right. My questions for my readers are:

1. Do you think this problem is appropriate for middle schoolers? (I do, though maybe not quite this early in the year.) They (the originating website) have it listed as an “Open Challenge” which they seem to be rating as harder than their elementary, middle and high school problems.
2. What hints would you give if they were stumped by it?
3. How many different approaches to solving this problem can you think of?
1. September 9, 2007 5:16 pm

You know, I forgot this one… I think I will use it, not because I think it is particularly interesting, but because it will be a way to introduce, almost violently “use smaller numbers / solve a smaller problem”

So my hint? 12345 is hard. Solve it for 12. Then 123. Then…

Instead of telling them directly to record their work, I will ask the incredibly annoying “how do you know you didn’t punch the same number into the calculator twice?” or just as bad “how do you know you didn’t skip one number?”

If you can con them into listing the numbers on paper, then, even if they don’t have a great leap of insight, you can help them find patterns in the list.

And once they free the digits from the numbers (but retain place values), then there are many ways to associate, commute, factor, etc, which is nice for the look back at the end.

Am I close?

2. September 9, 2007 5:23 pm

“Solve a smaller problem” was the thing I was thinking of as a first hint as well.

I have two ways of solving this one. The first is to think about how many of each digit will appear in each column of the addition. The second is to think in terms of the average of the allowable numbers, and how many there are. I wonder if I could come up with an accessible proof that the average is what it seems like it must be…

As far as “conning” them into writing the numbers down, I would also want to say them something to suggest that it will be much harder to generalize if they don’t keep track of what they are doing. 🙂

Thanks for the comment, JD!

3. September 9, 2007 11:37 pm

after 12 + 21 = 33 there are a few observations possible, but things may be too small to see.

123
132
213
231
312
321

Do you add all 6 1’s (222), 2’s (444) and 3’s (666), or do you note that the sum of each column is 6*2? Depends on the kid. I would like, if possible, for different kids to find different variations.

A simple form of averaging might look like:

123 + 321 = 444,
213 + 312 = 444
132 + 231 = 444

And there has to be more.

4. September 10, 2007 7:44 am

I agree that it would be good if different kids found different ways. They usually do, even though I only have 8 or 9 in the group.

I was noting the sum of each column. I like your way of adding all the 1’s, 2’s and 3’s separately as well, though that will get trickier with larger numbers.

That’s a nice way to look at the averages too. I was again thinking column-by-column (if there are 12 hundreds, there is an average of 2 hundreds per number, same for tens and ones).

September 11, 2007 7:47 am

My two cents worth:
There are 5*4*3*2*1=120 possible five digit numbers. For each of these numbers each number is used 24 times in each of the five digit positions. The sum of these 120 numbers in each digit position is 24*15=360. The sum of the 120 numbers then can be shown as:
24*15(10^4 + 10^3 + 10^2 + 10^1 +10^0)=
360(11111)=3999960

This maybe too much for some grade levels to absorb.

6. September 11, 2007 8:24 am

Cecil, that was my first approach as well. I also imagined that instead of realizing that the 360 should be multiplied by 11111, students might just think about it in terms of the addition algorithm — add up the ones, get 360, write down the 0 and carry the “36”, etc…

My second approach was to observe that the average of the 120 possible such numbers must be 33333, so the sum is 120*33333

I’ll will be trying this out on my middle school problem-solving class at some point this year, and will report back on how it went.