## Sunday, September 30, 2007

### Products of Digits: Challenges for Everyone...

[You may also want to look at the preview of the interview with Alec Klein, author of A Class Apart, to be hosted on MathNotations. Alec has agreed to answer my questions about Stuyvesant HS in NYC, other specialized schools and gifted education.]

One never knows where the inspiration for a math challenge might come from. This one came from a book I devoured recently, entitled The Righteous Men by Sam Bourne. If you enjoyed the Da Vinci Code and solving the number puzzles and other codes embedded in the story, you will definitely enjoy this novel. One of the most important clues to unraveling the mystery in the story was stated as a riddle:
Just men we are, our number few
Describable in digits two
We're halved if these do multiply
If we few perish then all must die.

Ok, now some of you are going to ignore the math problem and run out to find this book, but for those who would like a more prosaic version of the math problem, here is our first conundrum:
Note: Students from middle school on can attempt some of these although the proofs are a reach.

(1) Determine a 2-digit positive integer the product of whose digits is one-half the integer.
Now, it won't take you long to find such a number (once you get past the elliptical phrasing), but that's just to whet your appetite. The real challenge begins:

(2) Prove that your answer to (1) is unique, i.e., there is only one solution to the problem.

Comment: We're looking for more than an exhaustive search through all ninety 2-digit numbers or a programmed solution. The key to this and all of the remaining questions is to find an approach to solving a single equation which has 2 or more variables whose domain is the set of positive integers. Students are usually not introduced to solving such equations but they appear frequently on SATs and Math Contests. Because we are looking only for positive integer solutions, a standard algebraic approach must be supplemented with arithmetic concepts and testing of several possibilities. Number theorists refer to these as Diophantine equations.

(3) How about 3-digit integers? If 12 were permitted as the hundreds' digit, then (12)96 would satisfy the problem since (12)(9)(6) = 648, which is one-half of 1296. Unfortunately we can't allow that, so your challenge is to prove that there are NO 3-digit numbers (meaning positive integers) the product of whose digits is one-half the number.

Comments: Again, you're confronted with an equation but this time there are 3 variables. One approach is to solve for h in terms of t and u (I'll let you guess the symbolism) but there surely must be numerous possibilities! Or are there...

(4) Since we couldn't find a 3-digit number with the property, we'll rework the ratio:
Determine a 3-digit number the product of whose digits is one-fourth the number.

(5) Of course, I can't let you off the hook that easily. Prove that your solution in (4) is unique!

Standard disclaimer: These results have not been independently verified. Translation: I devised problems (2)-(5) and therefore there could be errors. If you find other solutions, let me know!

Reminder: Don't forget to give Proper Attribution when using these original problems. See the instructions in the sidebar. I appreciate your understanding and acknowledgment.

Totally Clueless said...

Very nice, Dave. Once you figure out how to do the two digit problem, the others seem to use the exact same method, though with more grunge.

Question: Does the last line of the verse have any math attached to it also?

TC

Dave Marain said...

thanks, tc!
One can always find mathematical meaning in most any phrase but I don't believe that is the case here! You have to read the book to interpret its meaning.

jonathan said...

I got the tens digit down to 2 and the units I had 4 choices... Altogether 8 trial and errors, but I would have been happier if I'd been able to narrow it more.

I'll play with the rest later.

Eric Jablow said...

Jonathan, the 2-digit problem is pretty simple.

The two-digit number ab is really 10a+b, and so we have

10a+b = 2ab
10a = (2a-1)b

2a-1 is relatively prime to a, so 2a-1|10. Since 2a-1 is odd, 2a-1 = 1 or 5.

a=1 leads to 10 = b, which can't happen.

a=3 leads to 30=5b, b=6, and the number is 36.

Dave Marain said...

Wonderful explanation, Eric. Ah, the power of higher arithmetic (we call it number theory nowadays)...

The 3-digit problem is a bit more challenging but similar as tc pointed out. I'll post my solution soon if I don't see one, although it may not be as elegant as Eric's.

Stay tuned for a 'traditional' advanced algebra problem that seems out of place on this blog. I can't wait to read your reactions to it...

jonathan said...

Three digits:

2abc = 100a + 10b + c -->
c(2ab - 1) = 10(10a + b)

c is even, 5 divides 2ab - 1, so 2ab ends with 6... ab ends in 3 or 8...
13c 31c 79c 97c; 18c 81c 24c 42c
29c 92c 36c 63c 47c 74c 68c 86c.

That's 64 possibilities to check, too many for me.

2abc = 100a + 10b + c -->
2abc - 100a = 10b + c
2a(bc - 50) = 10b + c

bc g.t. 50 (barely helps)
5b + c/2 is even (since we have two factors of 2 on the left).
So again, barely helps, but if b is even, c is 4 or 8, and if b is odd, c is 2 or 6

Down to about 30 trial and errors. Still too many? What did I miss?
132 136 312 316 792 972 976 184 814 818 244 248 424 428 292 922 926 364 368 632 636 472 744 748 684 864 868

Totally_clueless said...

Hi Jonathan,

c(2ab - 1) = 10(10a + b)

also implies that 10a+b is odd, so b is odd, which reduces your choices by half to 8. For these, you can try to explicitly find c using the above equation and should get a single digit number for a valid answer. You will find none.

TC

Totally_clueless said...

On 2nd thought, does 10a+b have to be odd? It only has to be divisible by an odd number.

Anyway, if you try to explicitly find c, then many possibilities are easy to rule out (c > 10).

TC

Dave Marain said...

Some nice thinking going on here...
My approach was less theoretical (I was analyzing this while driving back home on Sunday (don't say it!), so I didn't have the luxury of writing it all down. This forced me to be less theoretical.

Solving: 2h(tu-50) = 10t+u. [Don't laugh -- I really did that mentally!]
This implies the original number must be even, so u must be even.

Solving for 2h:
2h = (10t+u)/(tu-50).
Therefore the original number must be divisible by tu-50 and the quotient must be even.

Since tu > = 50, I claim these conditions reduce the number of possibilities to FOUR!

t=7,u=8: 78 is divisible by (56-50) but the quotient is 13 (odd).

t=8,u=8: 88 is not divisible by (64-50).

t=9,u=6: 96 IS divisible by (54-50) but the quotient, 24, leads to h =12. This was that bogus solution I mentioned in the post.

Finally, t=9,u=8: 98 is not divisible by (72-50).

I believe we're done! Any errors?

Andrée said...

I just put up my solution. I included a sentence or two about middle school students and how they might attack this. But you have to click on the solution because I couldn't get it all from Equation Editor to the blog without it shrinking. I'll be working on fixing this but my computer is being awful tonight and I have to toss it out the window now.

You can read it here. Thank you.

Andrée said...

Fixed. Computer's life is on reprieve.

Dave Marain said...

andree (sorry for omitting accents!)--
I really appreciate your sharing this problem with your middle school students. Your proof was very nice and similar to the way I did it. It's interesting how many variations there are, each one depending on the FORM of the equation used for the analysis.

I am particularly impressed by the multiple representation approach your students have become accustomed to using. This is phenomenal! Using the power of Excel (particularly its graph capability) in a math class is well-known to some educators and fairly unknown to others.

My thoughts about proof at this level are that just as adolescents show extreme variations in physical characteristics in that age group, there is also a wide variation in their mathematical maturity. Some middle schoolers I've worked with (7th graders) are ready for proof and can even do them on their own after seeing a few models. For awhile I listened to 'experts' explain that a formal geometry course is inappropriate for a middle schooler because of immature brain development, meaning they are just not ready to handle the conceptualizations. Perhaps not many, but some can, and I'm not just referring to a few mathematical geniuses out there. I'll bet if you showed your students your proof and explained the main ideas, some would be able to follow and a few, if they could handle the algebra skill part, could do a similar one on their own. Hey, you never know until you try and I know that you are not afraid to take risks.

Using my problems/investigations with your students make writing these worthwhile. Thank you and let me know how it goes with that portfolio assignment.