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Too ambitious as a WarmUp for the 6th or 7th grader? Would they immediately employ the "Make it simpler and look for a pattern" strategy? Is the calculator appropriate for this activity? Is this really an activity/investigation?

Since I'm already regarded as an anachronism, I guess it wouldn't hurt to play word games here:

Posted by Dave Marain at 7:36 AM 4 comments

Labels: digit problems, investigations, middle school, patterns

As the school year is beginning...

Which would you conjecture is more likely:
No digits the same in a 3-digit number or no digits the same in a 2-digit number?

You have 30 seconds to choose one of these - - - - - - - - - -
NOW WRITE YOUR GUESS ON YOUR PAPER and compare with your partner. Take one minute to discuss your thoughts...

Alright, I know some of you take exception to wasting these 30 seconds. What could be gained from such 'blind' guessing without the time to really think it through and work it out. I often used device this to encourage youngsters to react instinctively and to learn to trust their intuition. How many times have all of us had the experience of not trusting ourselves, only to find later that we were right. If it turns out that this gut reaction is not supported by the data, then the mathematical researcher (or the experimental mathematician in this case) revises the hypothesis. Ultimately, one attempts to validate one's conjectures via logic (deduction, induction, etc.). If you're still not convinced this is worthwhile, it's only a suggestion...

Now we're past the prelims. Our goal is to have our students begin with solving a particular case of the problem above and then to develop a general relationship for:
The probability that an N-digit positive integer will have N different digits. Of course, N is restricted to be in the range 1..10. We would hope our students from middle school on would recognize that the probability for N = 1 is 100%, whereas the probability for N = 11,12,13,... would be zero! Yes, we would hope!

(1) Show that the probability a 2-digit positive integer has different digits is 90%.
Comments: This is a well-known and fairly basic problem, but this is just the jumping-off point for this investigation. Various methods are likely here, depending on the background of the student. The middle school student (and many secondary as well) would likely list or count the number of 2-digit numbers with different digits. Some would realize that it might be easier to count those with identical digits and subtract from the total. More advanced students may use more sophisticated approaches for this and the other parts below. One could use this activity to develop the multiplication principle, permutations, use of factorials, etc. However, there is much to be gained from 'first principles.' Careful counting and making an organized list never go out of style!

(2) Show that the probability a 3-digit positive integer has 3 different digits is 72%.

(3) Complete the following table up to N = 10:
Note: P(N...) denotes the probability of the indicated outcome.

Number of Digits N.....P(N different digits)

...........1......................100% or 1
...........2..................... 90% or 0.9
...........3......................72% or 0.72
...........4......................50.4% or 0.504......
.
.
..........10.....................................................


(4) Time to revisit your original conjecture.... Explain why the probabilities decrease as the number of digits increase.
Note: One could give a purely descriptive explanation here.

(5) For more advanced groups:
Develop a formula for P(N).

(6) For more advanced groups:
Enter your expression from (4) into Y₁ of the Y= menu in your graphing calculator. Set up a TABLE with Start value of 1, increment (Δ) = 1 and Auto for Indpnt and Depend. Display your table and check the values you found from your own table.

(7) [Optional]
Closure: Write 3 ideas, methods, strategies, mathematical principles, etc., you have learned from this activity.

Posted by Dave Marain at 4:33 PM 3 comments

Labels: counting problems, digit problems, discrete math, investigations, probability, writing in math

The title question would be a medium level math contest problem.

The following could be the SAT version:

369121518...99
The integer above is formed from consecutive positive integer multiples of 3 from 3 to 99 inclusive. What would be the 50th digit in this number?

A student could theoretically take the time to list all the digits up to the 50th, however, this would be more time consuming than using logic.

Possible approach for SAT-type:

369 uses 3 digits.
This leaves 50-3 = 47 more digits.
Starting with 121518... there are two digits for each multiple of 3, so we divide 47 by 2, producing 23 with remainder 1. We now need to determine the 23rd multiple of 3 beginning with 12:

12 + (23-1)3 = 78. [At what point should students know this formula for arithmetic sequences?]

The remainder of 1 means we need to look at the first digit of the multiple of 3 that comes after 78: 81
So the answer would be 8, but, of course, I could have erred in my logic or calculation, so pls check that!

One could now assign the title problem for extra credit or as an extension.

Comment: How many readers believe that middle schoolers on up tend to rely on their calculator to find remainders. This has been discussed previously on this blog and a couple of algorithms were given for computing the remainder from the decimal result given by the calculator.

Posted by Dave Marain at 4:38 PM 13 comments

Labels: digit problems, logic, math contest problems, SAT-type problems

[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.

Posted by Dave Marain at 7:40 PM 12 comments

Labels: digit problems, middle school math, number theory

The 'answers' to the problems below are now posted in the comments...

[Update: I modified today's problems for my 9th grade group. An excerpt from the handout is shown at the bottom. How do you think they did? These are youngsters who need very precise instructions and find math more challenging. Guess how they did?]

How many 4-digit numbers have exactly 2 identical digits?

Today's questions were inspired by a Google search from one of the viewers of this blog. No matter how many of these types our students may attempt, there is no substitute for a systematic approach and clear thinking. Professional mathematicians who have done numerous problems of this type and are therefore likely to know an efficient method still can make careless logic errors. Imagine how our students do! They're looking for a quick and dirty approach, one short-cut method or formula that covers all kinds. Not likely!!

There's an interesting semantics problem with this question: Does 3232 count as a possible answer? Your first instinct might be to say, "Of course, it doesn't count!" But how many identical digits does it really have? For this reason, I will reword today's question to:

How many 4-digit positive integers have exactly one pair of identical digits?

[Note: Someone will probably argue about the semantics here as well but I'll let it stand!]
I'll post the result I obtained later but, for now, how many different approaches do you think your students could come up with? I thought of at least 3 not to mention writing a short program to count them!]

I see this kind of problem as appropriate for middle school and high school. For the middle grades, students should begin with the 3-digit version (see worksheet below) and be encouraged to make an organized list. By high school, they should be able to move on to more powerful counting methods but we know some are stuck at the 4th grade level of counting in an unorganized manner. In fact, from my observations, most high schoolers start in the 1000's, count those and multiply by 9. This is actually a fine method but should they know other approaches by the time they complete Algebra 2?

The following is a portion of the worksheet I gave to my group today. It worked out well. Any thoughts? Notice that I modified the 4-digit problem to make it more accessible for them. It provided an easier extension.
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Posted by Dave Marain at 5:37 AM 8 comments

Labels: combinatorial math, digit problems, SAT-type problems