Friday, December 14, 2007

What is the Largest 3-digit Multiple of 7? - A Middle School Activity on Remainders, Multiples and the Division Algorithm

Please don't forget to give proper attribution when using this activity in the classroom (see sidebar).

What is the largest 3-digit multiple of 7?

Would you expect most students to reach for the calculator and, without much thinking, test 999,998,997, etc., until they reach 994?

This is an opportunity to review the Division Algorithm, the conceptual meaning of remainder and how important this integer can be when solving problems involving multiples, factors, and various number theory questions (not to mention those repeating pattern problems so popular on standardized tests).

From the calculator, students obtain a result like 1000/7 = 142.8571429. The last digit is rounded which disguises the repeating decimal but some will know that. You ask for the remainder from this division problem and you may get an answer like 6/7 or blank stares or some decimal. Because we live in a calculator environment, students may have already acquired ways to obtain the remainder from the calculator display. Older students with their sophisticated graphing calculators may have a function that returns the remainder (like mod) or some application they downloaded for this purpose.

Here's one way students find remainders on their calculator:

Ignore the decimal, that is, look at the greatest integer value of the quotient, namely 142. Then, the remainder can be obtained from: 6 = 1000 - 142⋅7. This is just another form of how students should check their division:
142⋅7 + 6 = 1000

Students often think about this procedurally, not conceptually. In fact, a thorough understanding of remainders and the division algorithm would make the questions in this post fairly simple.

In general, suppose B is divided by A, producing an integer 'quotient' Q and a remainder R, where R is an integer satisfying A > R ≥ 0. Then the Division Algorithm states:
QA + R = B or R = B - QA

As an aside, some students are taught or discover another calculator approach for finding the remainder:
Subtract (discard) the integer part of the decimal, leaving 0.8571429, then multiply this result by the original divisor 7. Like magic, the display reads 6. Students should realize that the calculator internally stores more places than it displays! It may be worth it to demonstrate why this method works or have students investigate this, so I'll make it part of today's challenge:

Explain why the above procedure works for finding remainders. You may want to employ an algebraic derivation, using A, B, Q and R as above.

Students should now use remainder concepts to solve the original problem and the following:

What is the largest 6-digit multiple of 7? Explain your method carefully.

Note: If time permits (like right before a vacation), you may want to introduce modulo arithmetic (congruences, etc.) to solve the remainder problem in a more compact form:
10 ≡ 3 (mod 7) → 106 ≡ 36 ≡ (32)3 ≡ 23 ≡ 1 (mod 7), using the fact that 9 is congruent to 2 modulo 7. Thus, the remainder is 1 and the result follows. Of course, there's a great amount of overhead in developing this much number theory, but if you're looking a holiday challenge...


Anonymous said...

It must be the season... 3 days ago I did two remainder problems with most of my classes (I blogged, if you are curious).

Your algebraic verification of how to use a standard calculator to obtain remainders is very nice. Really, lots more kids should be exposed to it.

Finally, your feed no longer shows up in my Google reader - I don't know when you write any more. Did you know that?

Have a wonderful holiday.


mathmom said...

I am also having a problem with Dave's feed (but not his comments feed, ironically).

Jackie said...

Me too. I just realized I still get comments in my feed, but no posts!

Dave Marain said...

Happy Holidays!
I re-directed mu Posts feed through Feedburner a couple of weeks ago and there have been some complications. You may need to re-subscribe. I'm not sure about Google Reader -- I'll check it. Also, I've set up a new icon in the sidebar to allow readers to subscribe to the comments feed via Feeburner also.

BTW, are you or your readers able to post comments via email in WordPress? I'm sure this is easy to do in Blogger but I haven't seen instructions yet - my ignorance, you know!

IMHO, remainders are very important and all students need to have a better understanding of this both with and without the calculator! That's why I published this post.

Dave Marain said...

Uh oh! Did I mess up this blog's post feed for all of my regular readers? You may need to re-subscribe using the icon in the sidebar (the smaller one). The larger one is the Comments feed which seems to be working. I apologize for this incovenience - I became enamored of Feedburner and I've been tinkering - a very dangerous thing to do apparently!

Could each of you indicate which feed service you normally use? If Google Reader is in the majority, I may just go back to that, but, right now, I'm concerned about making more changes!

In spite of all that, HAPPY HOLIDAYS!

mathmom said...

I use Google Reader, and it was pointing to

I'll try adding the feedburner feed. But I'm surprised that the default blogger feed stopped working just because you added a feedburner feed.

Jackie said...

I too use Google Reader. I subscribed to the new feed, seems to be working fine now.