[Pri5 20170426FEM] Baking Éclairs and Macaroons

Problem / Question

     This problem for primary 5 from one of my acquaintances on Facebook, considered to be of intermediate level difficulty (in Singapore).  But it looks rather challenging to draw all those bar diagrams, doesn’t it?

     Here is my quickie solution without explicit algebra and without bar diagrams.

Solution

     For convenience, we use 6 circle units for Eclairs and 6 square units for Macaroons.

Suppose there were half as many Eclairs and Macaroons, then there would be 15 more Eclairs.  So 3 square units add 15 can be changed to 3 circle units.

     Add 15 to the 17 and change 3 square units to 2 circle units.  We deduce that 5 circle units is the same as 85.  From here we can easily figure out the rest.

Ans: 102 éclairs.

Comment

     The problem can be solved by bar diagrams.  However, there are many ways to skin the cat.  For more good stuff, please join my Facebook group “Effective and Elegant Mathematics”.

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H06. Use before-after concept

H08. Make suppositions

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School / Elementary School Mathematics

* any precocious or independent learner who is interested

[Pri_20151223WNSV] Unravelling Four Whole Numbers

Problem

If  A,  B,  C  and  D  are whole numbers such that  A × B = 8,  B × C = 28, C × D = 63,  B × D = 36,  find the values of    A,  B,  C  and  D.

Introduction

     This question seems to be taken from a secondary school textbook from a chapter on linear equations.  However, I think a good  primary school pupil could attempt this.

Strategy

     The key to solving the above problem is to make observations.  When you multiply up the first two equations, you get an  A,  a  C  and two  Bs  in the product.  Hmmm ... This doesn’t look promising ...  Ah!  But when you multiply the second and the third equations together, you get an  B,  a  D  and two  Cs  in the product.  This can cancel (via division) with the fourth equation which has one B  and one  D  in the product.

Solution

Remark

     Always cancel as much as possible, to avoid large numbers and reduce chances of making careless mistakes.

     By the way, a whole number is a non-negative (zero or positive) integer that does not contain any fractional part.   As such, the set of whole numbers is {0, 1, 2, 3, 4, ...}.  Thus we do not need to consider the negative square roots.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics (Challenge)

* Lower Secondary School Mathematics (Challenge)

* other syllabuses that involve whole numbers

* anyone game for a challenge

[NumTh Expository] The Principle Behind Casting Out Nines

Introduction

Think of a number ... say 685932.  Divide by  9  and take the remainder.      685932 ¸ 9      = 76214 r 6  Add up the digits,  divide by  9  and take the remainder.      6+8+5+9+3+2 = 33 ¸ 9 = 3 r 6  What do you notice?  Try this with any other positive whole number.

Discussion

     Did you see that a number and its sum of digits always have the same remainder when divided by  9?  This is the principle behind the method of “casting out nines”, used in the past for checking arithmetical calculations.  Why does this work?  Where is its magic?

     The decimal number system that we use today is based on the number  10, which is just  1  larger than  9.  Observe that  9, 99, 999, 9 999, 99 999, ... etc are all divisible by  9.  Hence, the powers of 10, namely 10⁰ = 1,  10¹ = 10,  10² = 100,  10³ = 1 000,  10⁴ = 10 000,  10⁵ = 100 000,  etc  all leave a remainder of  1  when divided by  9.   Thus in our example,

     685932 = 6´10⁵ + 8´10⁴ + 5´10³ + 9´10² + 3´10¹ + 2´1

                  = 6´(99999+1) + 8´(9999+1) + 5´(999+1) + 9´(99+1) + 3´(9+1) + 2´1

                  = 6´99999+8´99995´999+9´993´9 + 6´1+8´1+5´1+9´1+3´1+2´1

                  = 9 ´ something + 6+8+5+9+3+2

As you can see, all the “´1” allow us to separate out the digits, and then the stuff with 9, 99, 999 etc can be lumped together as 9 ´ some whole number, but we do not need to care too much about this multiple of 9 as it would not make any difference to the remainder.  It is now obvious that  685932 and 6+8+5+9+3+2=33 will have the same number when divided by 9.

     Let us generalise the argument.   If two numbers  x  and  y  have the same remaider when divided by 9,  we say that  x  and  y  are congruent modulo 9, and we write     x  º  y  (mod 9).  Congruence is an equivalence relation and “º” behaves in many ways similar to “=”.

Theorem

For an arbitrary number  n   with digits  [dk...d3d2d1d0]                                         n º dk + ... + d3 + d2 + d1 + d0   (mod 9)

                 n = d_k ´10⁴ + ... + d₃´10³ + d₂´10² + d₁´10 + d₀.    

Since  10^k º 1 (mod 9)  for all integers  k > 0,  we have

                 n º d_k´1 + ... + d₃´1 + d₂´1 + d₁´1 + d₀

                 n º    d_k  + ... +   d₃   +    d₂  +   d₁    + d₀     (mod 9).   © (Q.E.D.)

As an example of application of this principle, please refer read thisarticle.

Suitable Levels

* Primary School Mathematics Olympiad

* syllabuses that involve congruences and Number Theory

* anybody who is interested

[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

[OlymPri20150527NTSD] Sudoku with Modular Arithmetic?

Question

Introduction

     This is a pretty nasty-looking challenge.  It is like Sudoku, but only worse.  You probably cannot avoid some degree of trial-and-error (or [H07] guess and check).  There is a solution posted somewhere, but it is very long and chases many dead-ends before arriving at the solution.

     So how do we avoid making so many wild guesses?  We do so by making good use of number theory, which includes modular arithmetic.  In what follows, I am going to use the theory and notations as in my article (“Divisibility Tests for 7 and other digits”).  Please read the article to remind yourself of the various divisibility tests.

Solution

     It is easy to see that  j = 0. 

     Since abcde ¸ 5,  e Î {0, 5}.  But  0  is already taken, so  e = 5.

     Since divisibility is transitive (x ¸ y  and  y ¸ z  Þ  x ¸ z),  anything divisible by  4,  6  and  8  is also divisible by 2.  So  b, d, f, h  are all the even digits  2, 4, 6, 8  in some order, and  a, c, g, i  are the odd digits  1, 3, 7, 9  in some order.  From the divisibility tests for  3,  6  and  9,

     a+b+c º 0 (mod 3)

     a+b+c+d+e+f º 0 (mod 6)               Þ   a+b+c+d+e+f º 0 (mod 3)

     a+b+c+d+e+f+g+h+i º 0 (mod 9)   Þ   a+b+c+d+e+f+g+h+i º 0 (mod 3)

we get

     a+b+c º d+e+f º g+h+i º 0 (mod 3)

To narrow down the search space, we always test the group of digits that contain the most clues and the least unknowns (if possible).  Currently, this seems to be the middle group d5f.

     Using modulo 3,  d+5+f º 0   Þ   d+f º 1   Þ   d+f = 7, 10, 13  (because  2+4 < d+f < 6+8)

Since  d  and  f  are even digits, we can cross out  7  and  13.  So {d, f} = {6, 4}.

     abcd ¸ 4   Þ   cd ¸ 4   Þ   10c + d º 2c + d º 0   (mod 4)

If  d = 4,  then  2c º 0   (mod 4)   Þ   c º 0   (mod 2).  Not possible, as  c  is odd.

Now we know  d = 6  and  f = 4.  The number is  abc654ghi0,  and  {b, h} = {2, 8}.

Can  h = 8?  If so, we would have  4g8 º 4´100+g´10+8 º 2g  º  0 (mod 8).  g  º  0 (mod 4) 

g ¸ 4 ¸ 2.  But this is not possible, since  g  is an odd number.  Hence  know  h = 2  and  b = 8.  The number is  a8c654g2i0.  Now

4g2 ¸ 8   Þ   4´100+g´10+2 º 2g+2 º 0 (mod 8)   Þ   g+1 º 0 (mod 4)   Þ   g º -1 (mod 4)

   Þ   g Î {3, 7}.

If  g = 3, then working in mod 7, we have  a8c6 º 543,  and then

     Þ   a´1000+8´100+c´10+6 º 5´100+4´10+3

     Þ                  -a+8´2+c´3+6 º 5´2+4´3+3

     Þ                          3c – a + 1 º 4

     Þ                        a – 3(c – 1) º 0 (mod 7)

Note that  a  is an odd digit  and   c  is odd,  c – 1  is even,  3(c – 1)  is even and so  a – 3(c – 1)  is odd.  a – 3(c – 1) ¹ 0  because 0 is even.  Since  1 < a < 9,  a – 3(c – 1) = 7.  We need  a = 7  and  c = 1.  This would mean  i = 9.  But   g+h+i  º 3+2+9 º 2 (mod 3).  It is supposed to be 0 (mod 3).  So  g ¹ 3  i.e.  g = 7.  The number is  a8c65472i0.

Going back to mod 7,     a8c6 º 547

                               3c – a + 1 º 1

                                            a  º 3c

(a, c) = (3, 1) or (9, 3).  But if  (a, c) = (9, 3),  i = 1.  g+h+i  º 7+2+1 º 1 (mod 3), which is incorrect.  So we must have  a = 3 and   c = 1.  The remaining digit is   i = 9. 

The mystery number is 3816547290.  Bingo!

Heuristics Used

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H07. Use guess and check

H08. Make suppositions

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Hxx*  At each step, try to attack the part that has most clues and the least unknowns.

Suitable Levels

* Primary School Mathematics

* anyone game for a challenge requiring modular arithmetic

[Pri20150523WNCA] Trees arranged in a Hexagon Outline

Question

Introduction

     This is real eeeasy peasy lemon squeezy, isn’t it?  54 ¸ 6 = 9  Ta da!  The answer, right? Wrong!  You got tricked!  Ha!  Ha!

     Always tryto understand the question and do the planning first.  Never be in a hurry and jump to thecalculation stage.  So what went wrong?  Well, the tree at each vertex is counted twice.
     Huh?
     Sometimes to understand the situation, it may be easier to consider a simpler problem.  Let us say there are four trees per side.  This is how it looks like from above.

     You can see that the corner trees (coloured in orange instead of brown) are counted twice, because they each serve as an extreme marker of two of the sides of the hexagon.  There are 18 trees and if you divide by  6,  you get  3  and not  4.  One way to count properly is to start from one corner tree and count groups of three trees, either in a clockwise or anti-clockwise (American: counter-clockwise) direction.

     Notice that the number of trees on one edge of the hexagon is equal to the number of trees in one group plus one (the corner tree for the next group).  So for  18  trees, the correct calculation is  18 ¸ 6 + 1 = 3 + 1 = 4  for the number of trees along one edge.  We use the same procedure for  54  trees.

Solution

     number of trees on each side = 54 ¸ 6 + 1 = 9 + 1 = 10

Final Remarks

     You may want to generalise it into a formula

                    # trees on each side = total # trees ¸ #sides + 1

However, I do not recommend that you purposely memorise this formula.  Mathematics is not about memorisation.  It is about understanding.  Once you understand it, the formula comes out automatically.  You may test yourself or get a friend to test your understanding by setting a similar question but changing the number of trees and number of sides.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers

[S2_20150523XFDS] Numbers that can be Difference of Squares

Question

Introduction

     This is likely an primary mathematics olympiad-type of question, but lower secondary pupils can also try this.  It involves deeper thinking.  But where do we begin?  Sometimes it is good to begin from the beginning, and then follow your nose. 

Reminders

Solution

     Suppose  N  is a whole number such that  1 < N < 1000  and  N  can be expressed as

                                        N = a² – b²  = (a – b)(a + b)

a difference of squares.  So  N  can be split as a product of two factors  (a + b)  and  (a – b).  Observe that     (a + b) – (a – b) = 2b,       which is an even number.

     The difference between the two factors is an even number.  This can only mean that the two factors are  both odd  or  both even.  You cannot have one of them odd and the other even, because when you subtract them, you would get an odd number.  We now have three cases:-

     Case 1a:  N  is even but not divisible by 4.

     Case 1b:  N  is divisible by 4 (and, of course, is even)

     Case 2:    N  is odd  i.e. both  (a + b)  and  (a – b)  are odd

Ans:  750

Remarks

     In the foregoing, it is possible for  b  to be zero.  0 happens to be a perfect square, because  0² = 0.  However, we need not worry about this, because the above algebra is general enough to cover the case where  b  is  0.
   We have solved the problem using logic, even-vs-odd analysis and the three important algebraic identities under reminders (highlighted in orange).  We also used the special cases (highlighted in light blue) and made observations based on them.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* Primary School Mathematics Olympiad
* Secondary 2 Mathematics » grade 8 (expansion and factorisation)
* anyone who is game for a challenge in algebra and number theory