[OlymPri20151105NTC9] An x to Find by Casting Out Nine

Question

In class, John was thinking of a 6-digit number, A.  He added up all the digits and got the result, B.  Then, he subtracted  B  from  A, which gave a result, another 6-digit number, whose digits consist of 0, 2, 4, 6, 8 and  x.  Find a possible answer for  x.

Introduction

     This problem looks mind-boggling.  There are so many possible 6-digit numbers and there seems to be no clue as to how to even begin.  This problem hinges on a forgotten fact that many people used to learn when electronic calculators were not so prevalent.

Old Wine Most Fine

     It is a fact that any number is equivalent to its sum ofdigits in the sense that they both have the same remainder when divided by 9.  (see this article)  This is the principle behind the method of “casting out nines”, used in the past for checking arithmetical calculations.  Mathematics is never out-dated.  In fact, some of the old forgotten theory may sometimes turn out surprisingly useful.  If two numbers  x  and  y  have the same remaider when divided by 9,  we can write  x  º  y  (mod 9)  but in this article, I shall just write  x º y.  If   x º 0,  it just means that  x  has no remainder when divided by  9  i.e.    x  is a multiple of  9.

Solution

     Since                             A º B      where  B = sum_of_digits(A),

                                    A – B º 0

             sum_of_digits(A – B) º 0

           0 + 2 + 4 + 6 + 8 + x  º 0

                                  20 + x  º 0  [ 20 + x  is a multiple of  9]

                                     \  x  º 7  [x  is a digit & the next higher multiple of  9  is 27]

Remark

     One possible value of  A  is  864738.  Then  A – B = 864738 – 36 = 864702

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Primary School Mathematics Olympiad

* syllabuses that involve congruences and Number Theory

* anybody who is interested

[Pri20150306MOD] A Telescope Too Far?

Question

Introduction

     This is a nasty multiple-choice question to be set for a primary 6 (~ grade 6) pupil for a test.  It could also be a time trap as the pupil might spend a lot of time to no avail just to try to score that miserable mark.  With advanced knowledge, we can solve this using telescoping sums (a.k.a. the method of differences), where many terms cancel and the sum can be shortened, much like a telescope.

     However, this solution looks like over-kill.  Is there a solution that is more accesible to a primary 6 pupil?  Read on!

A Simpler Solution?

     We observe that a denominator of  5  is common.  In fact, since every term has an even factor, we have a common factor of 2 in the denominator also.  That means every term in the sum can be expressed as ¹/₁₀ of something [ Heuristic H09 ].  We can quickly work out the first few partial sums and express each of them as  ¹/₁₀ of something.  [ Heuristic H04 ]

     Tabulating the results [ H02 & H03 ], we observe that each partial sum is  ¹/₁₀ of something slightly less than ½.  So we try to express them as ½ minus something.  How?  We can subtract to find out.  [ H05 ]  For example,  if  ⁵/₁₂ = ¹/₂ – ???   then   ??? = ¹/₂ – ⁵/₁₂ = ¹/₁₂.

     Later on, we notice   [ H04 again ] another pattern: the factors in the denominators of the subtracted quantity matches the last two denominators of the last term.  If we considered the full sum, then what would those factors be?  (See the part highlighted in yellow).  These would be 20 and 21, as per the last term of the series.  Using this pattern, we work out the required sum as follows

That is the answer!

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards
H09. Restate the problem in another way

[Pri20150302PBW] Pebbles on a Road Backwards

Question

A jar contained some pebbles. Nigel took out half of them plus 2 more for display. Jenny took out half of the remaining pebbles plus 1 more. Finally, Alice took out half of what remaining plus 5 more for her project. In the end, there were only 8 pebbles left. How many pebbles were in the jar at first?

Solution

We can trace what happens to the remaining pebbles at each stage, and represent the information given in a sequence as above.  Note that when “Nigel took out half of them plus 2 more for display”, we would have half and then two less the number of pebbles.  Similary for the Jenny’s and Alice’s stages.

                            

Now we just do the opposite of all the operations.  The inverse operation of ‘–5’ is ‘+5’, the inverse operation of ‘´ ½’ is ‘´ 2’ and so on.  Working our way backwards, we arrive at 112.

Ans: There were 112 pebbles at first. 

Commentary

     Remember: In mathematics, there is no such thing as "the only way" to solve problems.  There are many ways to skin the cat, as they say.  Other methods to solve this question include: using algebra, and “the branching method” (taught by some tutors/teachers) which tracks both the remaining pebbles as well as the ones taken away.

     If you do use algebra (of some sort), remember to use different letters e.g. v, w, ... etc to refer to different things.  Likewise, do not just write  ‘½’  to represent ½  a unit, otherwise it could be misinterpreted as literally the number ½.  I suggest you can surround the ‘½’ with different shapes (e.g. circle, square, triangle) .

Precision is very important in maths, as well as in life. You don't want an imprecise person to be your aircraft designer, accountant or doctor, do you?

Heuristics Used

H05. Work backwards

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and fractions

* any learner who is up for a challenge

JCCDQBHWH_FN021(b) Range of Composite Functions

[original source unknown]

Introduction

     This question is from the usual book which did not credit the source.  It comes from some unknown junior college in an unknown year.  It is a challenging question because most students are poor at finding the range of composite functions.  Furthermore, this question has a little twist: you are given the range of the composite function, but you are required to solve for something.  So you need to, in a way, work backwards and/or use inequalities (another weak point for many students).

     Students are reminded of the right-to-left convention for functions in JC as well as GCE ‘A’ level examinations.  This means in ‘fg’ the  g  is done first before the  f.  There are some university professors who use a left-to-right convention, but here we do not.  So take note.

     Again metacogntion and heuristics are very important and I will illustrate their use.

Stage 1:  Understanding the Problem

What is this (part of a) question about?

range of composite functions, solving for unknown

What is given in the question?

The range of  fg.

What is the question asking for?

The value of  k  that leads to the given range.

Stage 2:  Planning the strategy

What heuristics do you think can be used for this question?

· Working forwards (considering the meanings, asking “so what?” “what next?”)

· Setting up equation/inequality

· Working/thinking backwards

· Consider equivalent expressions or rephrasing the problem

Can you recall the definition of the range of a function?  The range of a composite function?

Stage 3:  Execution

Any observations that can make your job simpler?

     Yes.  Observe that  g(x)  is a quadratic with positive  x²  coefficient.  So this is a parabola that looks like a happy smile.  To locate the minimum point, we can complete the square (a technique learnt in secondary school).

     g(x)  =  x² + 2x – 1  =  x² + 2x + 1² – 1² – 1  =  (x + 1) ² – 2

when  x = -1,  g(x) = -2.  The minimum point is (-1,-2).  So the range of  g  is all the

numbers from  -2  upwards.  i.e.  R_g = [-2,` \oo `).

So what now?

     With the two-stage method, suppose now we have

                  x ` \in ` R_g

That means?

                  x > -2

That means?

                  x + k + 1 > -2 + k + 1

Why do you do that?

     I want to slowly manipulate the LHS to get  ln(x + k + 1)  which is  f(x).  Continuing,

                  ln(x + k + 1) > ln(k – 1)

                                f(x) > ln(k – 1)

i.e.                            R_fg = [ln(k – 1), ` \oo `).

Why is there no switching in the inequality sign?

     The slope of the graph of  the natural logarithm is always positive (albeit getting less steep for increasing  x).  So applying  ‘ln’  on both sides does not change the inequality.

What is the clue again?

     We are told that  R_fg = [ln 3, ` \oo `).  Aha!  *epiphany*  *light bulbs flashing*

ln(k – 1) must be equal to ln 3!!!  This can be solved easily!

figure 1 – working forward and backwards

Stage 4:  Evaluation

Is the answer correct?

     Substituting  k = 4,  we see that    ln(x + k + 1) =  ln(x + 5)  and with  x > -2,  this will be  > ln 3  as given in the clue.

And why  x > -2?

     This is because g(x) > -2, which we knew  from completing the square.  We treat the  ‘g(x)’  as the  ‘x’  when applying  f,  because this is what  fg(x)  really means.

Stage 5:  Reflection

What did we learn from solving this question?

     We used metacognition to do self-monitor and self-questioning during the 5 stage problem-solving process.

     We used the following heuristics.

· Working forwards (considering the meanings, asking “so what?” “what next?”)

· Setting up equation/inequality

· Working/thinking backwards

· Consider equivalent expressions or rephrasing the problem

     We learned to apply the definition of the range of fg.  There are two possible methods: the one-stage method and the two-stage method.  The latter is usually better.

     In the two-stage method, the range of  fg  is found by first finding the range of  g  and then applying the function  f  to it.  After the first step of finding the inequality for  g(x),  we can simply use  x  in the formula for  f.  How?  We set  x  to be in the range of  g(x) from the previous step,  then slowly manipulate the inequality until the expression for  f(x)  appears.  This will give us the range of  fg.

     From the formula for the range of  fg,  we learned how to make use of the given clue to work backwards to find the unknown k.

     Difficult questions can be tackled by thinking systematically and logically, and using heuristics and metacognition.  Mathematics is hard, but it is fun after you have learned it.  If you have really learned it, you become more powerful because you can use the same technique to solve all kinds of problems in future.

Any of your own reflections?  Please post in the comments below.