Math Challenge: All-digits Arithmetic

Dennis Coble, @DennisCoble, just tweeted me this challenge an hour ago.

Here’s 1 that might interest you. Numbers 1-9 all used: 3 digit number, plus or minus 3 digit number, gives another 3 digit number.

As usual, the problem is deceptively simple, as is the solution. However, students could be engaged in their activity of this task for a full period. πŸ˜‰ And the ordering of student ability and success could be shaken. πŸ™‚

Enjoy.

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18 thoughts on “Math Challenge: All-digits Arithmetic

  1. Dennis informs me that he has two answers, and solutions, to this challenge. I have none so far. He suggests that he never was good at math. I claim that I am. Heh, heh, heh.

    Oh, it’s on!

    I invited him to guest post about this challenge since he offered it. I hope he accepts.

    In addition, I am thinking about designing a math lab around it. The Math I am encountering is fairly interesting. Perhaps one of my later posts will be a demo of me working through my activity of this task and designing a lab around it.

    Any thoughts on how to enrich this task to enrich the activity?

  2. For the solutions I came up with, I first started with the solution for the 3×3 grid that we called the magic square. Using the numbers 1 through 9, and filling in the grid, each row, each column, and the 2 diagonal rows of 3, all total 15, when they are added.

    • As usual, comments to my posts often continue in Twitter. This is what Dennis and I tweeted a couple of minutes ago.

      Ah. @rctoypalacecom is spilling the beans on my last math challenge http://bit.ly/kRRGKT . : ) Is it a challenge if you are given hints? πŸ˜‰

      @stefras It still doesn’t solve it completely. Some might not even know about magic squares, or solving them πŸ™‚ If you wish, I can delete it

      @rctoypalacecom Nah. Keep it there. I always like teasing students. Why have them if you can’t tease them, right?

      Nasty! πŸ™‚

      Have fun solving the challenge, folks.

      Note: Someone stole Dennis’ twitter account, so he had to change from @rctoypalacecom to @DennisCoble. Since the tweets were written as given here, I decided not to change the tag. But, please, direct your queries to Dennis to @DennisCoble.

    • As the post suggests, dumb luck was likely the best way to find an answer, but now we have to ask why is the answer correct and are there others, like Dennis’s answer.

      We look now for patterns.

  3. Sorry for first comment re solution. Saw the puzzle and answered without realising method was the goal, not the solution. Mine was based on an assumption with a minor tweak. I will know better in future. Mea culpa.

    • Not at all Ross. Answers and solutions (methods, Math) are both welcome. Thank you for yours.

      Answers basically tell us we are on the right trail, but the method tells us how to travel. The method informs and connects the Math and the answer.

  4. Further to the assumption method… I basically used a 3×3 Sudoku grid to test number placement. Numbers fell in place nicely. Didn’t check for other solutions.

    • This is basically how Dennis approached the problem. He started with a magic square, which is a special case of the 3×3 Sudoku grid.

      Is this a necessary method, folks? Or an arbitrary one? Does the math hold and reveal an underlying Math?

  5. We are getting great answers and explanations of these answers on Twitter; patterns are emerging using the 3×3 Sudoku grid (magic square) as a starting point.

    Thank you

    Before I post these, I would like to thank and congratulate the following people who so far supplied correct answers to this challenge.

    1. Dennis Coble who so far supplied the problem and two answers,
    2. Ross Mannell who supplied a third answer, and
    3. Earl Samuelson who supplied a fourth.

    Update: Someone stole Dennis’ twitter account, so he had to change from @rctoypalacecom to @DennisCoble. Since the tweets were written as given here, I decided not to change the tag. But, please, direct your queries to Dennis to @DennisCoble.

    The tweets, though chronological under each heading, are topically arranged, so the conversation thread is broken.

    Definitions

    Defining lending, carrying and borrowing

    @stefras: @rctoypalacecom Sorry. I am adding, rather than subtracting. In addition, you lend. In subtraction, you borrow. #mathchat #edchat

    @rctoypalacecom: @stefras Ok, that is the same as I learned as ‘carry’. The digits beyond 10 when doing sums, are carried to the next column #mathchat #edchat

    So an addition can lend or carry forward a digit. This digit is lent or carried. This same digit is borrowed back in the corresponding subtractions.

    Defining answer, family and variation

    @stefras: @rctoypalacecom Yep. So, for each addition, you have two subtractions. Technically you also have two additions, but commutative. #mathchat

    @stefras: @rctoypalacecom If you have a subtraction answer, you have two corresponding addition answers as well. So, you have 6 ans. #mathchat #edchat

    The six answers comes from Dennis’s two distinct answers and the three siblings (defined below) each of them forms.

    @rctoypalacecom: @stefras I have 2 subtraction solutions, which are also addition solutions, when the total is added to the subtracted amount. #mathchat

    @stefras: @rctoypalacecom Are your two subtractions related, ie. just rearrangements of the subtrahend and difference? #mathchat

    @stefras: @rctoypalacecom So now you have three “base” answers (here defining each +/- “family” as one answer)? #mathchat

    Here we are aware of three of the four distinct or “base” answers found so far. Each set of additions and subtractions that correspond to each other form a family; each one of these operations is then a sibling, as explained below and distinguished from cousins.

    @stefras: @rctoypalacecom Rearrangement may occur if there is no lending. Order of columns is irrelevant then. So you might have many answers!!!!

    @stefras: @rctoypalacecom Let’s call those cousins, since they are rearrangements of columns. Siblings: rearrangements of rows. #mathchat #edchat

    @rctoypalacecom: @stefras I have 1 family myself. Ross’s family was different. In my family, I have 2 variations, and Ross’s also has 2 variations. #mathchat

    Here cousin and variation are synonymous, referring to the same answer with different column arrangements. Sibling and family (of siblings) refer to the same answer with different row arrangements; this includes switching operation from subtraction to addition. So, any given family has four siblings: two additions and two subtractions. Cousins and siblings are rearrangements of the same answer.

    Later, we define parent as the sum or minuend in the family.

    There is also a rotation variation, defined below.

    @rctoypalacecom: @stefras The 4th one I found is a 90 degree rotation from the first solution sent out. The rows become columns. Like ‘magic’, LOL

    @stefras: @rctoypalacecom I was testing that. @RossMannell’s idea of using Sudoku and yours of using magic square make rotation likely.

    Turns out this was a good property to test. There is also the possibility of different combinations of flips and slides producing related variations of the same answer. These are being tested.

    Emerging patterns

    How many lends?

    @rctoypalacecom: @stefras In theory, then we have 2 families, and each family has capacity to shift, but only from 1s to 100s column. #mathchat

    @stefras: @rctoypalacecom That depends on the number of columns that are lent carried values. In @RossMannell’s answer, 8+5=13, so 2 arrangements possible. #mathchat

    @rctoypalacecom: @stefras those I tweeted are my solutions. Ross’s solution works the same way. The 1s column can shift to the 100s column, and work. #Mathchat

    @stefras: @earlsamuelson @rctoypalacecom I think the pattern here is one column (not the hundreds one) must carry or lend to another and the other two must not.

    @rctoypalacecom: @stefras In my solution, that is true, only one column carries over.

    So far, we think only one column of digits adds to a double-digit sum. One of the others adds to a single-digit sum even with a digit lent to it. These two columns must remain adjacent to each other, lent-to column left of lending column. The hundreds column cannot add to a double-digit sum; this would produce a four-digit answer, violating the problem’s constraints. Additionally, the third column must (?) also add to a single-digit answer and not be lent to. This column occurs to the right (ones column) or left (hundreds column) of the other pair.

    This set of assumptions must be tested for exceptions of one-lends that break the pattern and two-lends (no-lends has already been tested out; there are no viable no-lend answers).

    Do only certain digit-pairs lend?

    @stefras: @rctoypalacecom So far, the digits that add to create a double digit sum produce odd double digit sums. Again, hmm.

    Is it necessary that all double-digit sums are odd? Or are some even? Why? There are 16 possible additions that total to a double-digit sum; if only odd double-digit sums are utilized, only 10 of the 16 possible additions are viable in this problem. Again, are the others excluded, and why?

    Are 18, 27 and 45 necessary?

    @rctoypalacecom: @stefras I have to throw in another solution, based on new evidence. Every 3 digit combo that adds up to 18 is a parent.

    Again, a parent is the sum or minuend of a family. We think if the digits in this sum or minuend add to 18, the addition or subtraction is viable as an answer to the problem. I am testing why. If this is true, parent is a good title, since the “parent property” governs the family possibilities.

    @stefras: @rctoypalacecom Hmm. That is interesting. I wonder if it holds and why it is occurring.

    @stefras: @rctoypalacecom I noticed that the 3 digit addends @earlsamuelson and you found add up to 15 and 12; @RossMannell’s add 13 and 14; 27!

    @rctoypalacecom: @stefras Like when I mentioned the 18 total was in all 3. 45-18=27

    @rctoypalacecom: @stefras @vikkp The answers work because 18 is the first even number after 15, that is divisible by 3

    @stefras: @rctoypalacecom Spoiler! πŸ™‚ Back to a new drawing board.

    Fifteen is relevant since it is the magic number of a 3×3 Sudoku grid. We actually need to test whether 18 as a sum holds and explain why it is a necessary property.

    Other variations?

    @rctoypalacecom: @stefras The 4th one I found is a 90 degree rotation from the first solution sent out. The rows become columns. Like ‘magic’, LOL

    @stefras: @rctoypalacecom I was testing that. @RossMannell’s idea of using Sudoku and yours of using magic square make rotation likely.

    This rotation works for one (Ross’s) of the three non-rotated answers we have so far. It does not work for the other two. We have to figure out why, determine if this is necessary and perhaps find a connection between why and the fact that only Ross’s addends sum to 13 and 14, instead of 12 and 15.

    Last remarks

    @stefras: @rctoypalacecom @earlsamuelson @RossMannell each supplied a distinct answer to http://bit.ly/kRRGKT. Anyone else want to pitch in? #mathchat

    @stefras: @rctoypalacecom @earlsamuelson @RossMannell So now three distinct answers. Have we exhausted the list?

  6. Hi Shawn,
    Thanks for your message on Twitter and I too look forward to learning and sharing with you – we seem to have many common interests! This Maths thread is great, and I would like to use some of these ideas in my year 8 Maths class.
    I also teach a VCE (last year of high school) Environmental Science Online class. Our current area of study is Pollution, including bioaccummulation, endocrine disruptors and a case study of fluoride. My class blog is at http://vceenviroscience.edublogs.org. You may like to join us on Elluminate for a session some time?
    Best Regards, Britt Gow

    • Britt,

      Hi. I am always up for more learning. I will check your site.

      Thank you for your compliment on my posts. I try my best to deliver quality content. In fact, I am still learning most of the media tech, so much of my blog is experimental in nature. Several posts have not made it, since I slaughtered them in the making.

      Feel free to use whatever educational content I have on my blog and website. I created them for that purpose. I would appreciate an attribution, so I can retain ownership, and notification, so I can learn how the material works and helps in a classroom.

      Seriously, use and enjoy. I don’t create lesson plans that I can use in class, since I am a substitute teacher, so the only way these challenges and lesson materials that I create can be tested is through other teachers. Comments, critiques and suggestions (even requests) are encouraged.

      I’m thinking this particular challenge might be rather advanced, but then that really depends on what one asks for and how one does it. Kids can be very inventive and ingenious when they want to be, so I wouldn’t put any topic above any kid.

      Thank you for visiting and commenting. I am glad you found my blog enjoyable and useful.

      Have a great Winter (or is this season Summer down under and south of the equator? I could never get that straight.),
      Shawn

    • Hi Ty,

      Thank you for commenting. I will see what I can do and perhaps create a new post. I assume each digit, except 3, occurs only once in the question.

      I have a feeling though that Dennis Coble and Vijay Krishnan studied the question using programming <shake head>cheaters</shake head> on July 1 and they determined there was no answer.

      Shawn

    • Hi Ty,

      A found three quick answers to your puzzle, none of which have all digits unique. This was not a condition you placed on the question, but one which I added.

      Rewriting the question

      Given my understanding, stated in my last comment, your initial question

      • abcd *3 = efghi

      can be rewritten as:

      • a(a+1)(a+2)(3a+3) *3 = efghi, where a+(a+1)+(a+2) = 3a+3,
      • (a+2)(a+1)a(3a+3) *3 = efghi,
         
      • (a+1)a(a+2)(3a+3) *3 = efghi,
      • (a+2)a(a+1)(3a+3) *3 = efghi,
         
      • a(a+2)(a+1)(3a+3) *3 = efghi or
      • (a+1)(a+2)a(3a+3) *3 = efghi.

      Interpreting the question

      One can list the digits 0-9 as possible values of digit placeholders a-i, one of which, if each digit in the question were unique, is not used in the question. If any digit is used more than once, more than one of 0-9 is not used.

      One notices from the rewritten question that (3a+3) = 3(a+1), so d = 3a, 3b or 3c.

      Only [1,3], [2,6] and [3,9] are possible digit sets for [(a+1),(3a+3)]. So possible value combinations of [a,(a+1),(a+2),(3a+3)] are:

      • [0,1,2,3],
      • [1,2,3,6] and
      • [2,3,4,9].

      Further interpretation of the question reveals that if a is the first digit of abcd, it must be ≥4 if no lending occurs, ≥3 if 1 is lent from b*3, ≥2 if 2 is lent, et cetera, to make efghi a five-digit number.

      Solving the question

      To begin with, one can eliminate [0,1,2,3] as a possibility.

      • 0123 *3 = _ _369 (low extreme, 3 repeated)
      • 2103 *3 = _6309 (high extreme, 0 and 3 repeated)

      And [1,2,3,6] as another.

      • 1236 *3 = _3708 (low extreme, 3 repeated)
      • 3216 *3 = _9648 (high extreme, 6 repeated)

      This leaves [2,3,4,9] with the following answers:

      • 2349 *3 = _7047 (low extreme, 4 and 7 repeated)
      • 4329 *3 = 12987 (high extreme, 2 and 9 repeated)

      I’ll let you find the other two answers I found (so far). πŸ˜‰

      Remember you have to eliminate other possibilities to show you have found all answers to the question.

      Good luck and have fun.
      Shawn

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