[Pri20150303LIH] Lollipop in the Haystack

Question

The number of lollipops in a box is between  60  and  100.  If they are put into packets of 3, there will be 1 lollipop left.  If they are put into packets of 5, there will be 1 lollipop left.  If they are put into packets of 7, there will be no lollipop left.  How many lollipops are there in the box?

Introduction

     This question seems to involve writing lists of numbers and finding the elusive common number.  Some school teacher did just that starting from 3, 4, 5, 6 and 7, when the question already mentions that the number is between  60  and  100.  To add insult to injury, the boy who got this wrong in a test copied this teacher’s “model solution” as corrections!  Sigh!

     Is there a simple way to solve this mathematics problem without rummaging through the entire haystack, as it were?  The answer is: thankfully yes!

Solution

Ans: 91 lollipops

Explanation

First, I draw a table to analyse the remainders when divided by 3, 5 and 7.  [using heuristics H02 and H03].  The question asks for a number with remainders (1, 1, 0).  However, we can SHIFT the problem.  [H09]  If we consider one less than the required number, the remainders are (0, 0, 6).  We look for a number with such a profile.

Having zero remainders under division by 3 and by 5, this number must be a multiple of 15.  Trying 15 [H07], I get  (0, 0, 1) because 15 = 2´14+1.  To get a remainder-profile of (0, 0, 6), I multiply by 6, and I get 15´6 = 90. Now I just SHIFT back 1 to get the required answer!

List of Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H07. Use guess and check

H09. Restate the problem in another way

For Your Information

     In solving this question, I did not really use any Chinese Remainder Theorem or advanced mathematics.  I merely used heuristics that can be understood by most people, including the parents helping them and the teachers marking the test scripts.  J

The kids?  Oh!  They will be fine.  They will learn well if we equip them with powerful thinking skills but do not interfere too much.  Shift happens!   J

Related problem here.

[Pri20150303SJM] Race to the Bottom

Question

Thinking / Planning

Noting that Sarah spent all her money in both scenarios and that 22´(18´something) = 18´(22´something) , I am going to first guess that Sarah takes 18 days and 22 days to spend all her money in the 1^st scenario and the 2^nd scenario respectively. [ Heuristics: H07. Use guess and check & H08. Make suppositions ]  Then I try to adjust my educated guess.

Johari spent more money in the second scenario.  The difference in his spending among the two cases is 6, so we need

                10´(number of days₂) – 12´(number of days₁) = 6

Solution

If Sarah takes 18 days and 22 days respectively to spend all her money, then

the difference in Johari’s spending among the two cases is

                10´(22) – 12´(18) = 4

´³/₂:         10´(33) – 12´(27) = 6

Ans (a): Sarah has $22´27 = $594.
Ans (b): Johary has $(21+12´27) = $345.

Commentary

If the amounts of money involved were in the trillions, we would have thought that Sarah and Johari are certain countries, wouldn't we?

Anyway, a parent from a Facebook parent-support group posed this question asking for a simple solution without using ratios.  This question seems to be the equivalent of a system of simultaneous equations in four variables.  A few of us tried various methods to solve it, but all were quite complicated.  I knew this can be solved using algebra, but struggled for some time to give a simple solution.

I guess all solutions (even the one above) would have some notion of “ratio” hidden in it.  The reason is: every time you multiply or divide by something, there is actually a ratio involved.  But I hope this solution is “easy” enough.

[MathEd] Heuristics in Mathematics

Heuristics are guidelines or rules-of-thumb for doing some task or solving a problem.  They work most of time, but are not meant to be hard-and-fast rules.  It something does not work, try another approach.

The Singapore Mathematics Curriculum includes 13 heuristics (of which 11 are for primary school) that teachers are supposed to train their pupils for use in mathematical problem solving.  For convenient reference, this article lists these heuristics and some others, most of which are from my personal learning and teaching experiences (and a few by searching the Internet).  Making observations, connecting facts and using heuristics to solve problems is part of a good mathematics education.  Actually, there is nothing to stop a primary school pupil from using the more advanced heuristics, it is just that pupils should at least be taught the 11 listed.

Primary School (Elementary School)

H01. Act it out

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

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

Secondary School and beyond

H12 Think of a related problem

H13 Use Equation / write a Mathematical Sentence  

Other / Related Heuristics

· Look for clues, make observations & connections

· Considering the meanings / definitions

· Draw construction lines

· Identify important variables, use effective notation.

· Identify variable to eliminate

· Keeping one variable constant and observing changes

· Compare what you have currently with what you want (goal)

· Compare and Contrast Data

· Consider extreme cases.  Consider special cases.

· Generalise.  Solve by solving a more “difficult” problem.

· Work Forward (~H01)

· Check for plausibility (~H07)

· Formulate an equivalent problem (~H09, H012)

· Try to simplify & cancel.  Avoid complications. (~H10)

· Create something out of nothing 无中生有 (~H12, H10) e.g. x  ®  x+5–5

· Replace something by its equivalent 偷梁换柱  (~H12, H10) e.g. x ® e^ln x

· Recognise Chunks (Chunking), substitution (~H10)

· Check for parity (e.g. + or - sign, even or odd). (~H10, H11)

· Divide and Conquer, Divide into cases, Break set (~H11)

· Exploit symmetry (~H11)

· Remove denominator from complicated fractions 釜底抽薪 (~H11)
· Think of the opposite problem (~H12)

· Argue by contradiction (suppose the opposite is true)

[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

[Pri20150302MCP] Multiple Choice Questions with Penalty

Question

Solution

The ratio of the number of words spelt correctly to that of words spelt wrongly is 3 : 1.  I use triangle to surround the ratio units.  Multiplying by 10 and 4 respectively, we obtain 30 and 4 units respectively.  The score accrued is 30 – 4 = 26 units which corresponds to 130.  Dividing by 26, we get 1 triangle unit corresponding to 5.  And that is the answer required because it 1 triangle unit represents the number of words spelt wrongly.

Ans: 5

Check: 15´10 – 4´5 = 150 – 20 = 130.   Yes!  Woo hoo!!

Reflection

This question is similar to this fish and chicken cost problem, but there are no monetary units (e.g. dollars or cents) involved.  Also, because of the penalty for wrongly spelt words, we subtract instead of add.  We need to be mindful of the context of the question and to apply the appropriate arithmetic operation.

[Pri20150302FCW] The Cost of Fish and Chicken

Question

Solution

The ratio of the cost of fish to that of chicken is 1.5 : 1 or 3 : 2.  I choose this nice whole number ratio and write them in triangle units.  You can choose draw any shape you want to surround the ratio unit actually.  Multiplying by 3 and by 4 (the numerical dollar costs), we obtain 17 triangle units equal to 76.50.  From here we can divide by 17 to get the equivalent of 1 triangle unit, and then multiply by 3 to get the equivalent of 3 triangle units.

Ans: 1 kg of fish costs $12.50

Commentary

Once again my Ratio Units method solves this problem quickly and you can use this method to also present the answer in a matter that even the teacher who is marking this can understand J.  

[Pri20150302BGP] Boys, Girls and Party

Question

A group of pupils met at a party. Each child exchange phone number with everyone else. Melissa exchange phone number with 5 times as many boys as girls. Peter exchange phone numbers with 4 times as many boys as girls. How many boys and girls were at the party?

Solution

In the set up stage, we read the question carefully and transfer the information into a good representation.  I am using my Distinguished Ratio Units method i.e. using different shapes for different types of ratio units.We set up as follows:  One row for boys (B) and one row for girls.  Reserve one column for ‘original’.  In the first scenario, if you subtract 1 (Melissa) from the girls, you get 1 unit (let’s use a circle to mark it) and then the boys would be 5 circle units.  In the second scenario, if you subtract 1 (Peter) from the boys, you would get 4 units (let’s use triangle) for the boys as compared to 1 triangle unit for the girls.

Going back to the ‘original’ column, we use circle 5 to represent the number of boys (as this was unchanged in the first scenario), and we use triangle 1 to represent the number of girls (as this was unchanged in the 2^nd scenario).

With the set-up done, now my plan is to try to match up one of the types of units (either circle or triangle units).  I choose to match up the triangle units.  Take the relation indicated in yellow for Melissa that says ‘1 triangle unit minus 1 gives 1 circle unit’.  When we multiply this by 4, we get ‘4 triangle unit minus 4 gives 4 circle units’ and then we link this up the the existing triangle 4 (highlighted in yellow on the right).  We see that from circle 5, when you subtract 1 and then subtract 4, you get circle 4. 

This means that 1 circle unit is 5.  Adding 1, we see that 1 triangle unit is 6.  5 circle units makes 25.  The total number being 5 circle units and 1 triangle unit, we add up 6+25 and get 31, which is the answer. 


Commentary

In this word "problem", you start of with two original quantities and when you add/subtract something from one of the quantities, you get a certain ratio, and when you add/subtract from the other quantity you end up with another ratio.  This is a typical difficult type of question that stumps many students and many parents who trying to help them.

Many word "problems" in Singapore's Primary School Mathematics curriculum are just linear simultaneous equations in two variables in disguise.  The pupils will revisit this sort of maths problems in their secondary school years under algebra.  For primary school, simple algebra is taught but not emphasised.  Singapore is famous for her "Bar Model Method".  This method is good for visualisation for pupils at the lower primary levels.  However, at primaries 5 and 6 (~ roughly equivalent to grades 5 and 6), the numbers get bigger and the types of problems get harder.  You often have to cut or draw many bars.  It is especially challenging to draw bars to a suitable estimated length and if you draw the bars wrongly, you often have to erase the whole thing and redraw.  In their year 6, pupils take the Primary School Leaving Examination (PSLE), a high-stakes examination that determines the secondary school and stream (track) that they will go to. .  Many anxious parents and tutors cope by teaching the kids algebra.  However, algebra seems to be socially frowned upon and used only as a last resort.

Last year, reflecting on the above-mentioned challenges, and upon realising that many people use the same letter "u" (for unit) or same bar to represent different types of units and end up getting confused, I cooked up my Distinguished Ratio Units method.  It is powerful and avoids the need to draw and redraw the diagram, hence allowing the pupil to concentrate on the thinking and solving, rather than wasting time with drawing.  I admit that my method is actually algebra in disguise.  The bridging tactic where you equalise one unit to allow the other type of unit to connect is like substitution and elimination.  One triangle unit and one circle unit, for example, can be interpreted as variables x and y, preparing the pupils for secondary school.  The triangle and circle units also link with the concept of ratios which they have learnt in their middle primary years.  By the way, you do not need to use triangles and circles.  You can use squares, diamonds, heart shape etc.  Who says you cannot be creative in mathematics?

I hope you like my new method.

Here is a related problem.