Puzzle – quick sum
December 7, 2007 pm31 5:20 pm
Discuss the following question:
Using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 exactly once each, create a group of 1 and 2 digit numbers whose sum is 100.
For example, 47 + 30 + 12 + 5 + 6 + 8 + 9 = 117, a bit too much
My professor shared this with me yesterday. He credits Polya for a discussion, but can’t find the reference. Perhaps it’s from a talk.
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JD2718 
Does this even have a solution?
Consider this:
We have S=0+1+2…+9 = 45.
Let us take off a & b from the sum, and add 10a+b to it (replacing two one digit numbers by a 2 digit number), and assume that the sum is 100.
Thus we have 45-a-b+10a+b = 100
Thus 9a=55, which means a is not an integer
It seems to similarly follow for having more 2-digit numbers also.
Thus, my (possibly clueless!) conclusion is that there is no solution.
I came up with the same “solution” as “Clueless”.
We start with the sum of the single digits which is 45. We want the one’s digit to be 0, so we either need to remove the 5 or a combination of digits that sums to 15, 25, 35, etc. from the one’s place and “promote” it/them to the 10’s place.
If we just promote the 5, we end up with a total of 90 (50 + 1 + 2 + 3 +4 + 6 + 7 + 8 + 9, for example). If we try to promote more than one number, with a sum of 15, we end up with a sum over 150.
That’s how it looks to me: any number of promotions and demotions is going to leave you with a sum that is divisible by 9. 100 is not divisible by 9, so we can’t reach 100.
Another way to see it is to remember that old trick that tells you a number is divisible by 9 iff the sum of it’s digits is divisible by 9. You can generalize this a bit to say that
Or the remainder mod 9 of a two digit number equals the sum of the remainders of the digits. So no matter how we construct one and two digit numbers out of the digits, we can’t change the residue mod 9 of the sum.
Of course, I make a lot of mistakes so I may be clueless too.
Rolfe’s reasoning is the most elegant. Great job!
So here’s the follow up: if I give this to a student, should I at some point tell the student to stop working?
Polya pointed out that many would continue to work well past the point that a real problem solver would have started proving it impossible.
And yes, the answers from Clueless to Dorn are good, and I like Rolfe’s, too.
Maybe a hint will help: Ask the students to observe if there is anything common to the sums they get with repeated experiments, and then draw a conclusion (they should observe that every sum they get is a multiple of 9). The hint may be given after a few minutes of work. With this, I think they will arrive at the answer.
Rolfe, I love the divisibility by 9 angle.
JD, interesting question about how long to let kids “mess with” it. I don’t have a good answer/intuition on that.
Your posts (puzzles) seem to be very popular, even too much popular, because they are solved after a few hours already, so I can’t come up with my solution unless I am surfing the web all day long ;) Of course, I could avoid looking at comments…
Edgar, sorry about that. I often make a separate place for answers; I will do this more.
So here goes. Maybe we just throw it out there, and wait until someone says “maybe it can’t be done” and then challenge them to demonstrate that clearly. Maybe it’s not such a bad thing to have a question that goes unanswered now and then.
What do you think?
No, no, it’s nothing, I’m just so talking. Everyone still can do the thinking of his own without looking at comments, can’t he? So, you can do as you wish, of course!
That is a good question. Here are some of my (conflicting) thoughts.
(1) It is good to let students struggle for a while on problems like this so they’ll remember in the future that sometimes the struggle is hopeless and a different approach is needed. If you tell them right away that there may not be an answer, then you’re not teaching them to always consider that possibility. You’re only teaching them to consider it when warned.
(2) Problems I’ve come across in my work (as a programmer of one sort or another) often don’t have answers, and it is never very comfortable to conclude that something can’t be done. Especially when you’re talking about things like security. Even when I was solving this little puzzle I hesitated — I was sure that I couldn’t solve the problem using the little framework I’d put together, but I hadn’t taken the time to convince myself that I’d covered every possibility. Now it’s pretty obvious to me, but it never is at first. I think that giving kids some practice making conclusions like that is a good thing. That sort of doubt is something I never had in school.
Also, I’ve had plenty of problems go unanswered but I keep thinking about them. It wouldn’t be a bad thing to let that happen in class, especially if you try to come back to it every now and then.
(3) If you present a problem as solvable when you know it is not, some of your students might be pretty PO’d at the dishonesty. So you might want to present it carefully, asking “Can you do this?” instead of saying “Do it!” Or make sure they’ve seen you run into similar situations first.
So how non-committal is this? I think that it’s good to let them see you solve problems like this, it’s good to let them struggle then help them out, and it’s good to leave some questions unanswered. You just need a lot of questions.
So we’ve established that the sum of any such construction must be divisible by 9.
What’s the minimum number that you can construct?
And the maximum?
And can you construct _all_ multiples of nine in between?
1. Pseudonym has some nice extensions. If we hold the restriction to 1- and 2-digit numbers, you’ll pardon me for solving “What is the minimum?” “What is the maximum?” and posing the “Can all in-between numbers be constructed?”
2. Rolfe, in my classes students understand that not every question has an answer. Some of the biggest differences between my class and a ‘typical’ algebra class are 1. harder factoring, 2. lots of fractions, 3. recognizing that “no solution” is a real answer, and 4. off topic problem solving that is far more involved than the ‘normal’ day-to-day algorithmic work.
Yesterday’s warm-up for my algebra refresher started with asking for x in: 5(3x – 2) – 4(4x + 1) = 14 – x.
I have, from time to time, left challenges on the board, even as we are doing regular work. Sometimes I offer a biggish number to factor (could be prime, could be 3599 ‘ish).
In that spirit I left this as a challenge on the side of the board. The teacher who shares the room liked it, and left it as well. Now there is a small buzz as students are asking each other if they got it. And I heard something interesting. A 9th grader approached a 9th grader, they were from the same middle school, and reminded his friend that their teacher had posed the same problem last year, but didn’t think they ever went over a solution.
I think I will leave this to ripen for a few days, and then devote a bit of class-time.
And to return to Rolfe, I contend that my students will, if and when they learn what’s going on, agree that the question was difficult, and that there had been some misdirection, but not at the level of being unfair.
It sounds like you have set good expectations for your students. The way you present the problem, I don’t think they would find it unfair at all.
I like the fact that you use a lot of fractions too. One of my big shocks getting to the ‘real world’ was that answers, when they were numbers, were almost never integers.
I heard this one on Car Talk and blogged it a long time ago.
My comm coll students have no problem solving it quickly, which was nice. The younger kids enjoyed it.