Edit: #4 below has been corrected. I am indebted to one of mathmom's astute students for catching my error!

Number theory is part of many states' standards but usually only at a basic level (factors, multiples, primes, composites, gcf, lcm). Below you will find a problem for your students to work on (preferably with partner). It is not an introductory problem using remainders so they would have needed to do preliminary work beforehand.

Here are some suggestions for developing the foundation for today's challenge problem:

(1 ) List the first 5 positive integers which leave a remainder of 1 when divided by 2? Describe, in general, such positive integers.

(2) List the first 5 positive integers which leave a remainder of 3 when divided by 13? If you subtract 3 from each of these, what do you notice? Explain!

(3) List the first 5 positive integers which leave a remainder of 12 when divided by 13. If you subtract 12 from each of these, what do you notice? If, instead you ADD 1 to each of the 5 positive integers, what do you notice? Explain!

(4) What is the least positive integer N, greater than 1, which leaves a remainder of 1 when divided by 2, 3, 4 or 5? [Ans: 61]

Note: The word 'or' may be confusing or inaccurate here. Modify as needed!

Now for today's challenge (allow use of calculator):

What is the least positive integer which satisfies ALL of the following:

leaves a remainder of 1 when divided by 2

leaves a remainder of 2 when divided by 3

leaves a remainder of 3 when divided by 4

leaves a remainder of 4 when divided by 5

leaves a remainder of 5 when divided by 6

leaves a remainder of 6 when divided by 7

leaves a remainder of 7 when divided by 8

leaves a remainder of 8 when divided by 9.

Notes/Comments

This challenge looks harder than it is. Variations of these often appear on math contests for middle school and beyond. Simpler versions like example (4) above have appeared on the SATs.

Of course, modular arithmetic and congruences would make this problem trivial but that is non-standard and requires more time to develop.

I will not yet post the answer or possible solution...

Here are some suggestions for developing the foundation for today's challenge problem:

(1 ) List the first 5 positive integers which leave a remainder of 1 when divided by 2? Describe, in general, such positive integers.

(2) List the first 5 positive integers which leave a remainder of 3 when divided by 13? If you subtract 3 from each of these, what do you notice? Explain!

(3) List the first 5 positive integers which leave a remainder of 12 when divided by 13. If you subtract 12 from each of these, what do you notice? If, instead you ADD 1 to each of the 5 positive integers, what do you notice? Explain!

(4) What is the least positive integer N, greater than 1, which leaves a remainder of 1 when divided by 2, 3, 4 or 5? [Ans: 61]

Note: The word 'or' may be confusing or inaccurate here. Modify as needed!

Now for today's challenge (allow use of calculator):

What is the least positive integer which satisfies ALL of the following:

leaves a remainder of 1 when divided by 2

leaves a remainder of 2 when divided by 3

leaves a remainder of 3 when divided by 4

leaves a remainder of 4 when divided by 5

leaves a remainder of 5 when divided by 6

leaves a remainder of 6 when divided by 7

leaves a remainder of 7 when divided by 8

leaves a remainder of 8 when divided by 9.

Notes/Comments

This challenge looks harder than it is. Variations of these often appear on math contests for middle school and beyond. Simpler versions like example (4) above have appeared on the SATs.

Of course, modular arithmetic and congruences would make this problem trivial but that is non-standard and requires more time to develop.

I will not yet post the answer or possible solution...

## 13 comments:

"This challenge looks harder than it is," he says. So I started with an index card for scratch paper. Hmph--it took me both sides!

I figured that the conditions for 2 and 4 were contained in 8, and the condition for 3 was contained in 9. I suspected that if 2 and 3 were satisfied, then 6 was automatic, but it took me a little while to convince myself that was true. I guess I haven't done enough of this sort of puzzle.

It's easy to find the numbers that satisfy both 8 and 5, and it's easy to test them for 9. Once I noticed that 9 is cyclic (should have known!), it was easy to jump to the next possible number. But testing for 7 is a nuisance.

I know a couple of elementary-age boys who will enjoy this. I'll pass it on to them as extra homework Friday.

I think that this problem yields quickly to a solution using LCM -- Dave how much do you want us to post or not in the comments at this point?

Densie or mathmom--

It's ok to post your answer/methods...

Actually, my bigger issue is how most middle school teachers would view this:

(1) too challenging -- only for the MathCounts kids

(2) May be accessible for most students if time is taken to develop the concepts of remainders, lcm, etc; however the curriculum does not allow for this time expense

(3) same as (2) but it's worth making time for

I am interested in knowing how much time middle school teachers spend on this and related topics in number theory for Grades 6-7-8. You can tell I'm advocating for some increased time for this and higher-order challenges but I'm also a realist about the pressures teachers face to prep kids for state-mandated tests.

Do you believe that the prelim questions I posed would be helpful or would they 'give it away?'

I'm also curious what you thought about the results of the Algebra 2 Exam which I posted the other day. Are you surprised by the poor showing or does it make sense given the current inconsistencies in foundation mathematics prior to algebra?

So, I figured that the desired number has to be one less than a multiple of 2,3,4,...,9. So the least such number should be 1 less than the LCM of these. (My guess is that is the leap that would be hardest for my middle schoolers to make.) Anyhow, the LCM is 2520 (2^3 x 3^2 x 5 x 7) and the desired number is thus 2519.

Dave, I think your lead-in questions do a good job of setting the stage for this.

I've done a series of problems like this with my middle schoolers in the past, but I demonstrated the method, rather than expecting them to find it. I like your lead-in questions and wonder if my kids could use them to find a method.

LCM is already in the curriculum, so giving problems that use it makes sense to me.

Mathmom,

People are lulled to sleep by long, monotonous-looking problems. Try the following with a student: ask him out loud "What is 8 × 4 × 3 × 5 × 0 × 6 × 2 × 1?" Usually, he starts working from the beginning of the question, not realizing that the answer must be 0.

It's the same thing here. You realized that the answer plus 1 must be divisible by all the integers between 2 and 9. That reasoning requires one to see the pattern, not just to start working the problem.

Always step back from a problem and look at it in its entirety. That can save you much time.

Thank you Denise, Mathmom and Eric...

Yes, the answer is 2519 (at least that's what I obtained!).

As you know my obsession is stretching minds of children, helping them to reason more deeply than is normally expected by the 'prescribed' curriculum. Guess I would call this the 'implied' curriculum. Number theory happens to be my passion so I naturally gravitate to it in the problems I write.

I truly believe that these kinds of challenges can be just as important for developing the mind and student understanding of mathematics as all of the real-world problems that we currently find in texts. But that's just me...

Duh! I didn't even think of the LCM. I tricked myself by thinking of division with remainder as multiple-plus-extra and not noticing it was multiple-minus-one. I think I'll go write 50 times, "I will use inverse operations. I will use inverse operations..."

Don't feel bad, Denise. Usually one of my students would see these things at lightning speed and feel sorry for me. I have worked with middle schoolers who just "see" it -- their brain can process it so much faster than mine. I would tell the rest of the class that we can all make sense of these ideas even if it takes us mere mortals a bit longer.

We teach because we want youngsters to experience these same revelations that we have had, each at their own time, when they're ready.

Without challenging the mind, it will not grow. Every morning I do my online word games (Jumble, crossword). I don't know why exactly but it keeps my brain from atrophying and, heck, I just enjoy it. Now you know why I write these math problems - it's as much for me as it is for my readers.

Eric, you can put that problem in writing, and many students will start working it left-to-right without noticing that the answer must be zero. There was a Math Olympiad (for grade 4-6) problem like that sometime in the past few years. The first problem is often some ugly-looking computation that has a cool trick to it. My kids may not always find the trick, but by now they know to look for one. :)

But that's just me...Dave,

I think you know at least your dedicated commenters well enough to know that it's not

justyou. ;-)Look at all this great stuff I missed while I wasn't reading.

This is wonderful -- easily accessible, but serves as entree to all sorts of deeper topics.

I amazed two classes with magic tricks on Friday, and this was at their core. I'll let the kiddies play with some basic congruence classes from time to time, and become more comfortable... maybe with this activity? I'll let you know.

Jonathan

I brought this to my middle school group today. This was largely successful, with about half the group getting through all the warm up and the challenge in our 45 minute math period. The rest will continue work on it on Friday.

One of my students discovered a "bug" in one of the warm up questions:

(4) What is the least positive integer N which leaves a remainder of 1 when divided by 2, 3, 4 or 5? [Ans: 61]

The answer to this question, as written, is actually 1. (So, on the spot, I told them to find the least positive integer N

greater than 1.) I had already modified the "or" part to read "...when divided by each of 2, 3, 4 and 5."I'll write more later about how the whole thing went.

Mathmom--

Congratulate that student for catching my careless error! Thanks as always for trying these out. You have an impressive group there - tell them I said so!!

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