Time for MATH MINUTE! (provide your favorite theme music here). 

         Get out your paper & pencil, because I have a new puzzle for you!!


Responses: PUZZLE #107 More about nines (revised)

 

       We saw some properties of nines in Puzzle 106, here are some more.*

 

A.    As in #106, take any 2 digit number ab, a and b not equal, reverse the digits, and subtract. The result is always a multiple of 9.

       Examples:     43-34 = 9,

                             83-38 = 45 = 5x9,

                             79-97 = -18 = -2x9.

       Explain why this works.

       Actually you can use any size number of digits (why?).

 

       This is like Puzzle 106 which was for larger numbers.

         Let ab be a 2 digit number with a & b different.

         Then if ba is larger than ab we have

         ba – ab = (10b + a) – (10a + b) = 10b – b – 10a + a = 9b – 9a = 9(b-a).

 

         You may have wondered why we didnt have a Mystery Number for 2 digit numbers?

         It turns out that these numbers do not reduce to a single number when you subtract them from their reverse: ab – ba is a multiple of 9, but it will determine another multiple of 9, and so on in a circle.

         Example: 72 - 27 = 45, 54 – 45 = 9, 90 – 09 = 81, 81 – 18 = 63, 63 – 36 = 27, 72 – 27 = 45, . . .

 

B.    Take any 2 digit number ab, add the digits together, a+b, and subtract the result from ab, ab-(a+b). The result is always a multiple of 9.

       Examples:     13-(1+3) = 13-4 = 9,

                             43-(4+3) = 43-7= 36 = 6x9.      

       Can you explain why this works?

 

       Using an argument similar to Part A., let ab be the 2 digit number, then

         ab – (a+b) = (10a + b) – (a+b) = 10a – a + b – b = 9a.

 

C.    Given any multiple of 9, the sum of the digits of the answer will always be a multiple of 9.

       Examples:     8x9 = 72, and 7+2 = 9,

                             121x9 = 1089, and 1+8+9 = 18 = 2x9.

       Why does this work?

 

       The key here is to notice that a multiple of 9 can always be written as a sum m of 10s and 9-m, so that the number m of 10s plus 9-m will equal 9 or a multiple of 9. For example 2x9 = 18 = 10 + (9-1), and 1 + (9-1) = 9.

 

***     Another way to look at this is in terms of the result in Part B. There we saw that the equation  ab – (a+b) = 9a holds. Now if ab is also a multiple of 9, say ab = 9n, then the equation would look like this: 9n – (a+b) = 9a. So if we solve for (a+b) we have (a+b) = 9n – 9a = 9(n-a), and (a+b) is a multiple of 9.

 

D.    The sum of the digits of a number is called the digital root of the number. The number can be any size. By Part C a number is a multiple of 9 if and only if the digital root is also a multiple of 9.

       Explain why this is true.

 

       By Part C we see that a multiple of 9 will satisfy the result that the digital root is also a multiple of 9.

         Now reverse the argument, and assume that the digital root is a multiple of 9. Then the number m of 10s and the number (9-m) will determine a multiple of 9.

 

***     Using the alternate argument in Part C, we have the digital root equal to a multiple of 9, and conversely, if it is a multiple of 9, say (a+b) = 9m, then by Part B we have ab – (a+b) = ab – 9m = 9a, so ab = 9a + 9m = 9(a+m), and ab is a multiple of 9.

 

E.    Casting out nines: This is a check (some call it a sanity check) on different arithmetic operations. It is not used so much with hand calculators these days.

       Problem: add up the column:       3264

                                                          8415

                                                          2946

                                                          3206

 

       First compute the excess of each number (divide the digital root of each number by 9 and list the remainder – this is the casting out operation), then add up the numbers and all of these excesses:

              3264 -> (3+2+6+4) = 9+6 ->       6

              8415 ->                                      0

              2946 ->                                      3

              3206 ->                                      2

            17831 ->                                    11

 

       Now reduce the answer and the sum of the excesses by 9:

       17831 -> (1+7+8+3+1) = 20 -> 2, and 11 -> 2.

 

       This tells us that the answer 17831 may be correct, in other words in order for the summation to be correct the two reduced sums must be equal. If the answers do not agree we know the answer is wrong!

             

       The technique also works for subtraction, multiplication, and division problems.**

       Explain why it works? (see **)

 

* See Wikipedia for more information on this:

     http://en.wikipedia.org/wiki/9_(number)

 

** http://en.wikipedia.org/wiki/Casting_out_nines

 

 

Have fun!

 


         Send your comments, ideas and solutions before Monday to the email below, and be sure to put * MM *  in the subject line:  parksjm@potsdam.edu

         Visit us here online at:  http://www2.potsdam.edu/parksjm/MM1.1.htm  to see the results every Friday/Monday.

         See you next time on MATH MINUTE!  (theme music fades out here).