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05 Oct 2005 - 22:40

## can someone check the answer?

Find all the numbers that satisfy all of the following conditions:

1. Positive whole numbers less than 100,
2. Four more than each number is a multiple of 6
3. The sum of the digits of each number is a multiple of 4.

### and what is the best way to do this problem?

We used Doug's number lines (WHICH ARE GOING TO BE GETTING A WORKOUT THIS YEAR, IT'S OBVIOUS). We labeled one number line with multiples of 4, and the other with multiples of 6. We didn't need the multiples of 6, but it made things easier to have all the multiples of 6 sitting there, where we could see them.

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## Comments

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Look here for syntax help.

looks like you'd wanna mark or just write down)
all the " 4 + 6n " numbers in the specified range
(i.e., {4, 10, 16, 22 ... 94} )
and then just inspect 'em for the "digit sum" property.

how you'd do it with a bigger upper bound
is another story; maybe you'd've gotten some
insight from doing the small problem, though ...

-- VlorbikDotCom - 05 Oct 2005

Hmm, the answer is ((3n+2)*6) - 4 (n = 0-4), but that was arrived at empirically. That formula works again for n = 6-9, and for n = 11-14. I can feel a trick inside of this one, and that it's related to the sum of the digits of multiples of 9, but I can't quite work it out.

FWIW, I worked the first few, then started looking for a pattern.

-- DougSundseth - 05 Oct 2005

nope, I got no insight

-- CatherineJohnson - 05 Oct 2005

We got:

8, 26, 44, 80

-- CatherineJohnson - 05 Oct 2005

4 + 6n

hmm

I'm thinking.

-- CatherineJohnson - 05 Oct 2005

This, by the way, is an excellent example of a Challenge Too Far.

I'm not at the point where this problem is going to teach me something.

I'm basically just doing a slightly more sophisticated version of Guess and Check, and I don't think Christopher would have done that much, even.

Left entirely to his own devices, he would probably just have started going through the numbers 1 to 100, one by one.

I prompted him to write out multiples of 6 on one of Doug's number lines, and multiples of 4 on one just below (we didn't make them proportional).

That exercise, though, was terrifically worthwhile, I felt. I pointed out, again, that skip-counting is the same as finding-all-the-multiples. And since we were going through each multiple of 4, Christopher ended up doing a Saxon Mental Math-type exercise, which was good.

-- CatherineJohnson - 05 Oct 2005

A much better problem--much better for me, and I may try it with Christopher to see if it's useful for him at all--would be this one:

1. Positive whole numbers less than 100, 2. Each number is a multiple of 4, 3. Four more than each number is a multiple of 6

This is what I had thought the question was, and I did it in my head.

The simpler form reinforced & developed number sense, which is always a good thing, as far as I'm concerned.

Tomorrow I'm going to give this one to Christopher, and see if he spots what to do.

-- CatherineJohnson - 05 Oct 2005

Russian Math is filled with highly challenging problems that are as challenging as they can possibly be without tipping over into student-regresses-to-guess-and-check.

-- CatherineJohnson - 05 Oct 2005

I don't get 4 + 6n

Still have to work through Doug's...

-- CatherineJohnson - 05 Oct 2005

"4 + 6n" was wrong ... sorry.

i shoulda said "6n - 4": #'s with the property
that when you add 4, you get a multiple of 6.
this is the famous "modular" (or "clock") arithmethic
at the heart of number theory:
since 6 - 4 = 2, we can refer to the same set
("all numbers of the form 6n -4") as the set of "2 + 6n"s:
search {2, 8, 14, 20 ... 98} for the digit-sum property.

but you've already solved the problem as stated;
again, sorry for the worse-than-useless "hint".

sundseth's formula (18n + 8) is intriguing --
seems to produce a number of the desired form
whenever n is not a (nonzero) multiple of 5 ...
i'll get back to you.

-- VlorbikDotCom - 06 Oct 2005

Catherine,

This answer is totally empirical: you cannot choose anything less than 10 because it doesn't have a sum of digits. (I don't think you can count that 8=8+0). Now, you can't choose any number that has an odd first digit because an odd plus an even digit is odd. So you are left with: 20, 22, 24, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 80, 82, 84, 86, 88. From here, I just did a brute force calculation on each number and came up with 26, 44, 62, and 80. I'm sure that there is a more elegant way once you narrow down the numbers.

-- AnneDwyer - 06 Oct 2005

"and what is the best way to do this problem?"

By computer.

for(int i=0; i<100; i++){

if(((i+4)%6) == 0){

int sum = int(i/10) + i % 10;

if((sum%4) == 0)printf("i=%d\n",i);

}

}

Too lazy to test it out. I hope I didn't screw it up.

The "%" is for modulo. A%B is the remainder of A/B.

I really don't like these problems for math. What fundamental skill or method are they trying to develop? Hopefully, the skill is to notice that some problems are much more easily done using a procedural computer language.

By the way, I've noticed the same thing in computer-based geometric modeling. Mathematicians are always looking for a unified mathematical formulation to topology rather than a procedural one. It hasn't worked yet.

-- SteveH - 06 Oct 2005

Darn, I lost my indentation!

-- SteveH - 06 Oct 2005

OK, here's what I got, as long as V wants to open up this can of worms, and as long as Catherine wants to know what the slick way to do it is.

The solutions k require:

1. k<100 (i.e., k is a two digit number)

2. k = 6n-4 for some n

3. k = 10*k1 + k2 for some k1 and k2 with k1+k2 = 4, 8, 12, or 16.

OK, any time you hear something about sums of digits, think about divisibility either by 3 or by 9. From the first relationship we get

6n - 4 mod 3
= -4 mod 3
= 2 mod 3.

If the sum of digits is a multiple of 4, then

10* k1 +k2 mod 3 =
(k1 + k2) mod 3 = 2 mod 3

(last step from the previous relation).

Now if k1+k2 is both a multiple of 4 less than 20, and equal to 2 mod 3 (i.e. of the form 3n+2 for some n), then k1+k2 must be equal to 8. It can't be equal to 4, 12, or 16, because none of those numbers is equal to 2 mod 3.

So we are looking at a subset of the two-digit numbers whose digits are both even (because the ones digit must be even hence both are) and which sum to 8. I guess we are in fact looking at all such numbers.

8 is okay because you can think of it as 08.

Modular arithmetic is, I think, the only slick way to do it. Since 6th graders don't know modular arithmetic, I assume guess-and-check was what was intended. I don't like this problem for 6th graders. It should be given to college students taking a first course in number theory.

-- CarolynJohnston - 06 Oct 2005

Bernie's comment on this problem was that it's a classic constructivist problem: it's all dorking around in the dark, and no satisfaction when you get the answer.

He also told me a wonderful story, apropos of this, that I'm going to blog on tomorrow.

-- CarolynJohnston - 06 Oct 2005

I actually think it isn't necessarily a guess and check problem for a 6th grader. Some tedious math perhaps, but it does provide some good multiplication practice. It also forces the student to develop an algorithm to solve the problem.

From the first fact you know you don't need to go past 100.

From the second fact, you can eliminate over 80% of those 100 possible answers by finding the multiples of 6. Then subtract 4.

And from the last fact, you just sum the digits from the answers in step 2 to get the final answer.

It's not elegant and maybe more trouble than it's worth (about 50 calculations), but at least it is multi-step, requires the student to approach the problem methodically, and practices some basic skills.

(x6)(-4)(sum)
6 2 --
12 8 8*
18 14 5
24 20 2
30 26 8*
36 32 5
42 38 11
48 44 8*
54 50 5
60 56 11
66 62 8*
72 68 14
78 74 11
84 80 8*
90 86 14
96 92 11
102 98 17

By the way, I never liked these types of problems.

-- KDeRosa - 06 Oct 2005

These are puzzles and - at the 6th grade level - not mathematics. By assigning a problem like this to a group of students, the constructivists are trying to develop generalized problem solving skills.

Such puzzles are motivating and enjoyable to some students, especially students with a competitive spirit. They have a minor place in the curriculum as time fillers. Let's say a substitute doesn't have a lesson plan and needs to engage the students in something useful. Let's say you have 15 minutes on a Friday afternnon in a period shortened by an assembly.

Such puzzles, if presented in a written form, can improve the very specific kinds of reading skills employed in story problems.

-- CharlesWilliams - 06 Oct 2005

Constructivist math is:

"all dorking around in the dark, and no satisfaction when you get the answer"

I love this.

There may be satisfaction that you got the right number, but you surely know that whatever you did probably can't be applied to anything else. That leaves kids with the sense that doing math is some vague process with no rules or methods beyond basic arithmetic.

"...the constructivists are trying to develop generalized problem solving skills."

... instead of real math skills. I would say that mastering real math skills and methods develops this "generalized" (whatever that is) skill much better. Constructivists don't believe this.

"Such puzzles are motivating and enjoyable to some students, especially students with a competitive spirit. They have a minor place in the curriculum as time fillers. Let's say a substitute doesn't have a lesson plan and needs to engage the students in something useful. Let's say you have 15 minutes on a Friday afternnon in a period shortened by an assembly. "

I agree.

-- SteveH - 06 Oct 2005

I am amazed how easy it is for a few people using little more than basic math knowledge to see right throught these so-called math problems -- they are barely math at all. Is it that our math eductors have so little math knowledge that they can't see this or just don't they care? I don't think I could live with myself if I had to teach with such a curriculum knowing how many children will be damaged by it.

-- KDeRosa - 06 Oct 2005

Of course, the truly sad thing is that this problem/puzzle would probably be the most difficult problem if it were included in the NAEP test.

-- KDeRosa - 06 Oct 2005

"Is it that our math eductors have so little math knowledge that they can't see this or just don't they care?"

I have been trying to figure this out for a long time. My guess is that those in charge of K-8 math did poorly in math in school, didn't learn any math in ed school, were very open to the agenda of the ed school pedagogues, and don't communicate with high school math teachers who are supposed to have degrees in math. If they don't support and implement what they were taught in school, then what do they have left? Nothing. We are basically saying that everything they learned in ed school is wrong.

What I find interesting is what I call a curriculum wall between K-8, which does not emphasize knowledge and skills, and high school, which does (at least for the college prep and honors tracks). I would like to hear from teachers about this problem. My feeling is that K-8 administrators really don't want the high school people telling them what to do - kind of a turf sort of thing.

For many kids, it's all over by high school. They are unprepared for the big change in content, skills, and expectations. People wonder why the academic gap is related to parent education. The parents (or tutors) make up the difference so the kids can make the big jump to high school. This curriculum wall also seems to keep the lower schools from tracking how their kids do in high school. Perhaps they don't want to know. Enough kids do well that it doesn't become a big issue.

-- SteveH - 06 Oct 2005

Liping Ma has writen that many elementary math teachers have poor understanding of the basic math they are teaching as compared to Chinese teachers, especially when it comes to fractions or, horrors, division with fractions.

This is one reason why I don't like when each school tries to remedy the deficiencies of the constructivist base curriculum they've chosen by "supplementing." Quite frankly, they are not qualified to do so.

-- KDeRosa - 06 Oct 2005

The sum of the digits equals 8 in all correct numbers. Therefore, all the possibilities are combinations of (0,8), (2,6), and (4,4).

-- SmartestTractor - 06 Oct 2005

Smartest Tractor:

You ARE smart.

I was groping for this as we did the problem; I knew there was some obvious, simple WAY.....and I just couldn't get to it, even with the number lines.

Though, in this instance, for this particular answer, number lines may get in the way.

Wonderful!

-- CatherineJohnson - 08 Oct 2005

"I was groping for this as we did the problem; I knew there was some obvious, simple WAY.....and I just couldn't get to it, even with the number lines."

This problem is neither obvious nor simple.

Its purpose is unclear.

And the work-benefit ratio is too high.

I give it an "F".

-- CarolynJohnston - 08 Oct 2005

wait a minute, tractor -- how do you know
that all the correct numbers have a digit sum
of 8 (until after having found 'em)?

i'm evidently an easier grader than carolyn and'll
gladly pass the darn thing (C-, say) --
part of the point i failed to make in bringing
"modular arithmetic" into the picture
might have been that it's easier (when you know how)
to compute the "next candidate" (next member of
{2, 8, 14, 20, 26 ...}) by simply adding 6
to the previous candidate (thus: 2 ... add 6: 8...
add 6: 14 ..." etc) than it is to compute 'em
"directly" ("1 times 6 is 6; subtract 4: get 2 ...
2 times 6 is 12; subtract 4: get 8 ... 3 times 6 ..." usw).

this "constant difference" trick
is sure worth knowing eventually.
i'm not willing to dismiss out-of-hand
the notion that one could profitably demonstrate it
at the middle school level (without necessarily
introducing the entire "modular artithmetic" apparatus).

also, with the "less than 100" condition removed
we have a problem i don't know how to answer--
and it seems like a pretty interesting problem ...
in particular, SteveH?'s computer "solution"
fails (even under the implicit assumption that
there's a C compiler handy wherever one happens to be).
base-ten "digit sums" are only friendly with 3 & 9;
i'll need an idea new to me if i'm ever to know the truth.

-- VlorbikDotCom - 08 Oct 2005

_this "constant difference" trick is sure worth knowing eventually. i'm not willing to dismiss out-of-hand the notion that one could profitably demonstrate it at the middle school level (without necessarily introducing the entire "modular artithmetic" apparatus)._

OK, OK. I'm in a grumpy mood this morning. I'll raise its grade to a D.

Actually in the 'modular arithmetic' treatment I posted above, I proved that its digits had to sum to 8. Maybe SmartestTractor was using that fact in her argument.

-- CarolynJohnston - 08 Oct 2005

Actually in the 'modular arithmetic' treatment I posted above, I proved that its digits had to sum to 8. Maybe SmartestTractor? was using that fact in her argument.

oh. yes. quite right. thanks. ... i have to admit
i'd quit following your calculations carefully
at right around the crucial point (which happens
also to be where you invoke the "2-digit" condition).

can i blame tractor (just a little bit)
for failing to cite a source? -- of course not!
this isn't exactly a scholarly journal or something!
just another instance of the oft=remarked-upon
"vlorbik can't read" phenomenon pure and simple.

hmm ... let's see ... i'll bet carolyn's "digit sum mod 3" trick
could be adapted somehow to the cases with more digits ...
(scratches head, wanders off in search of paper & pencil ...)

-- VlorbikDotCom - 08 Oct 2005

The problem stinks.

BUT, seeing as how I'm not at the Modular Arithmetic level, Smartest Tractor's answer was what I was trying, and failing, to come up with!

-- CatherineJohnson - 09 Oct 2005

I have to print this thread out and read it.....but SKIMMING, Charles is right; the point here is to develop 'general problem solving skills,' which we know, thanks to cognitive science, do not exist.

Problem solving skills emerge with the acquisition of beaucoup domain knowledge.

KDeRosa is also right that this isn't completely Guess and Check, although the assignment was sent home on a sheet titled:

##### Guess and Check

The first thing I had to do was quiz Christopher about whether he really, truly had to guess and check.

Although I wasn't going to come up with Smartest Tractor's strategy, I was going to come up with something (which I did, thanks to Doug's number lines.....and if Christopher was seriously supposed to guess numbers and check whether they worked, I was going to have to dump the number lines.

Finally I asked Christopher, 'Do you really have to guess and check, or can you guess and check in a smart way?' (Defining 'smart' broadly, that is.)

Christopher said, 'She said we don't have to totally guess and check,' or something like that.....so I decided we were going to give the number lines a workout.

He told me yesterday his teacher said it was OK that he used a number line.

-- CatherineJohnson - 09 Oct 2005

I doubt very much that his teacher has any slick tricks up her sleeve either.

In short, they do not know what they want a kid to get out of this problem. They don't have a good reason for giving it.

And for that reason alone it should be tossed out.

-- CarolynJohnston - 09 Oct 2005

WebLogForm
Title: can someone check the answer?
TopicType: WebLog
SubjectArea: MathProblemHelpLine, MiddleSchoolMath
LogDate: 200510051839