Week 1
Resources: 2nd sentence.
Week 2
Pg. 1 Dastardly diagrams
Pg 3 Dastardly diagrams 1st line
Pg 4 Last line
Pg 6 “A more straightforward...”
Week 3
Pg 1 line 2 of text
Pg 4 Q5
Pg 5, 1st paragraph “any guess you make...”
Pg 6 I’d replace “It doesn’t tell us what happens if we’ve a 6 on one side, or anything else for that matter. So if there’s a 7 on the other side of the S, we have some small confirmation that the rule is true (though this confirmation would be disputed by most logicians)” with “It doesn’t tell us that S should be on the other side only when 7 is on the first side – so we could have a 6 on the first side and an S on the other side without breaking the rule. ”
Week 4
Pg 1 last line
Pg 2 include “ the flea starts from the border of the table...”?
Pg3, 1st paragraph “two payment schemes” should be “three”
Pg 5, Answer 4, middle of paragraph “always the flea...”
Pg 6, Scheme 3, “your total in any given day is less than a power of 2” sounds a bit casual. Do all/most kids notice this by themselves? Aren’t they surprised? I’m a bit worried that they should be expected to accept this as a rule, without further checking/explanation – In particular as this is used as a rule in Week 5.
One intuitive check could be: ok, let’s see what happens when we add 1 to the total earnings in day n+1: 1+1+2+22+…+2n
Then 1+1=2 so you get 2+2+22+23+…+2n
Also 2+2=4=22 so you get 22+22+23+…+2n
and 22+22=2×22=23 so you get 23+23+…+2n...
so in fact, in the end you get 2n+2n=2×2n=2n+1.
On the other hand, in a similar situation my kid, who’s not so well versed in powers of 2, said something like “your total in any given day is twice the total in the previous day, plus 1”. Can we blame her? I think one could capitalise on these two different patterns with
Scheme 4: You’ll be paid 1 cent the first day, and in each of the following days you’ll get paid as much as in all the previous days altogether, plus 1 cent.
When done step by step this may bring about the proof above more naturally.
Week 5
Pg 4, 2nd paragraph I’d replace “you can’t head off from A towards C, and then, when you get to the centre, change direction and head towards D” with “you can’t head off from A towards D, and then, when you get to the centre, change direction and head towards C.
Week 6
Pg 4, 1st paragraph “you need to move” insert “ the disks”?
Week 7
Pg1, last paragraph, should it be 4×4×4?
Pg 5, 3rd paragraph “In order to see how to do this, consider the first swap. In this swap, R was the 1st letter and B the 2nd letter in the pair being swapped. (The goal of the swap is to rearrange this.) Arrange the letters so that the place marked R is free. “ -- In principle there are 5 ways to do this unless one guesses that in this case we want to place counter Y immediately after the free space. So the next sentence. ” This means that the counter R is now on the place marked B, and can slide across to the place marked R. The order (again, anticlockwise) of R and B has now been swapped” is not quite true, though it explains what the author had in mind in the first place. Wouldn’t be simpler to say that “in order to swap two adjacent counters, we rotate all counters until the two counters to be swapped are in the places marked B and G, while the place marked R is left free.”.....
Pg5 and 6, Solutions 1 and 2 start slightly inconsistently? If you’re allowed to leave one cup empty in 2, why shouldn’t you be allowed the same in case 1?
Also, the solutions seem intuitively clear but are not proven – should the kids try all possible cases and see that these are the best fits? Should they try a few cases and conclude there’s a pattern? In this case I’d suggest including a few preliminary cases in the statement of the Conundrum at pg. 3.
Pg6, line 4 Missing word in “if you’re not allowed...”
Pg 6, midpage, “There’s a nice Java applet...... ”
Week 8
Pg 2 I’d rewrite “The teacher deals 21 cards onto the table in 3 groups of 7” as “The teacher deals 21 cards face up onto the table in 3 columns of 7”
Pg3, middle paragraph: I’d find it easier to follow the argument visually, so I’d insert diagram with highlighted places showing where the middle column gets distributed when cards are dealt again
1 / 2 / 34 / 5 / 6
7 / 8 / 9
10 / 11 / 12
13 / 14 / 15
16 / 17 / 18
19 / 20 / 21
Week 9
Pg2 The Grinch paragraph switches from present tense to past tense
Pg 3 2nd paragraph “placing you finger”
Pg 6, Solution 1, I’d prefer the algebra, it’s shorter. The vertical times the horizontal back edge is 360cm2, and their ratio is 800720=109. Multiplying these two we get the square of the vertical side: 400cm2. So the vertical side is 20cm.
MC Handbook10
Change title from Week 12 to 10?
Pg 1 Resources: --Some schools don’t require a calculator with nCr button, Anna doesn’t have one. I’d skip this resource and have them multiply on paper, it won’t hurt them. (no calculators allowed in Olympiads).
--“A copy of the Question Sheet” I’d add in --“A copy of the COMMENCING COMBINATORICS Question Sheet” because for the hurried reader, it might not be clear that COMMENCING COMBINATORICS is meant to be activity 6. I know it’s silly but I stand by it.
Pg 1, Activities: you’d like to list their names to be consistent with the next Handbooks?
Pg 1, Activity 1, Magic squares, replace “finding total number of possible magic squares “ by “finding total number of filling in 3×3 squares with numbers 1 through 9 (most of them are not magic)”.
Pg 1, Activity 1 I’d add in reference to the week no when these were done.
Pg 1, Activity 2 I’d delete“(Do whole new list, don’t just add on to old list.) “ because:
Pg 4, Activity 2, 4 cards In view of the coming Activity 3, I’d list the pairs like this:
AQ / AK / AJQA / QK / QJ
KA / KQ / KJ
JA / JQ / JK
A row for each of the starting letters A, Q, K, J.
A column for each of the ending letters A, Q, K, J.
No pairs on the diagonal because we cannot pair a card with itself.
4×3 pairs in total.
Pg 4, Activity 2, 5 cards I’d just add one more row and column for 10.
AQ / AK / AJ / A10QA / QK / QJ / Q10
KA / KQ / KJ / K10
JA / JQ / JK / J10
10A / 10Q / 10K / 10J
Pg 4, Activity 3 I’d notice that each pair above the diagonal is now considered the same with its symmetric through the grey diagonal (as if we folded the paper along the grey diagonal). So we need only count the pairs above the diagonal: 1+2+3=6. As well, this is half of the initial pairs in the table: 1+2+3=6=4×32. Similarly for 6 cards.
Pg 5, Activity 4, 4 cards. To list all triples systematically, one could start from the pairs of 2 cards in the table above and try to place the letter A, Q, K or J in front, if possible (if that letter is not already in the pair).
AQK / AQJAKQ / AKJ
AJQ / AJK
KAQ / KAJ
KQA / KQJ
KJA / KJQ
QAK / QAJ
QKA / QKJ
QJA / QJK
JAQ / JAK
JQA / JQK
JKA / JKQ
4 tables of 6=3×2 triples each. Each colour represents all possible permutations for a choice of 3 letters. (In truth, these last four tables don’t achieve much for us now, so I’m not suggesting they should be included. I included them here because if one was doing the same with, e.g. 6C3 then one may start seeing the pattern 6C3=5C2+4C2+3C2+2C2 when discarding repetitions: in each table, we can discard the triples below the grey diagonal and above the grey horizontal line – just a note for later – possibly in the 2nd 3rd or 4th year J)
Pg 5, Activity 4, 2 decks: “can tell the different”...
Pg 7, last line should be ×35? Was there some confusion here as it’s not clear if the bonus number would be chosen from among the remaining ones or from among those already chosen?
MCHandbook 11
Title: Week 13 relabel
Pg 5, Q8: I found the questions immediately preceding the tables confusing. It’s a matter of which of two events you consider first . Take these two events:
(A) You pick any 6 no-s between 1 and 45.
(B) A set of 6 WN and 1 BN is drawn from no-s between 1 and 45.
“in how many ways each of the following can happen” can be interpreted in two ways:
O1) Suppose (A) happens, then (B). How many among the possible outcomes of (B) will match the result of (A)?
O2) Suppose (B) happens, then (A). How many among the possible outcomes of (A) will match the result of (B)?
In the solutions vii)-xii), on Pg9, it seems to me O2) is considered: the answers count possible outcomes of (A) once (B) has been decided. However, in the statement of Q8, “For the Lotto, you pick any 6 numbers between 1 and 45. At each draw 6 Winning Numbers are drawn out. Once these numbers are drawn out, a Bonus Number is drawn from the remaining 39 numbers. “ It seems to me that (A) is decided before (B), so I’d tend to go for option O1). Option O2) wouldn’t seem natural to me, since it assumes that once the Drawn has happened, I then blindly choose my 6 no-s.
It also seems to me that in the end we’d be interested in the probability of winning some Lotto prize under option O1)? (I don’t do Lotto) It does decrease our chances in comparison to O2), so it’d make sense for the state that Lotto be organized as in O1 J. E.g.
In case (vii), P(choose A, then B): P(choose B, then A)=6:39.
All in all, it seems interesting to compare O1 and O2, but on the other hand Q8 posed its own set of challenges even before getting into the O1/O2 discussion.
If you’d like to reformulate the question as O1 I think the answers would be
vii) 6C5x1C1x39C1
viii) 6C5x39x38
ix)6C4x2C1x39C2
x)6C4x39C2x37C1
xi)6C3x39C3x3C1
xii)6C3x39C3x36C1
They’re picked from the two sets: {no-s I have picked} and {no-s not picked by me} if I got this right.
In a sense O1) is counterintuitive: since it measures outcomes of (B), it’s as if we measure the chances of the state company, not of ours. But on the other hand it’s just a case of we made our choice first, it makes no sense to discuss what would have been had we chosen something else. This is reminiscent of, though in a sense different from, the over-used Monty-Hall problem:
(A) A car is hidden behind one of 3 doors, two goats behind the others.
(B) The guest chooses a door (call it door 1) and the presenter opens one of the other two (let’s call it door 2) to reveal a goat.
The guest should normally switch the door. My simplistic explanation is that people tend to confuse
O1) doing (A) then (B) with O2) doing (B) then (A).
O1) gives P(door 3)=2/3 whereas O2) gives P(door3)=1/2.
Indeed, in O1), (A) first says P(door 1)=1/3, P(door 2 or door 3)=2/3.
Then (B) second says P(door 2)=0 so P(door 2 or door 3)=P(door3)=2/3.
Pg5, (iv) The “E25,000 if you match the BN as well” I’d suggest falsifying the “win” rubric to “E25,000”, to avoid confusion.
Pg6, Q9 same comment about the formulation of the problem, but not a problem in this case, still I’d make it more precise.
Pg 6, Q9: “you choose 5 numbers, from 1 to 50 and 2 lucky Star numbers, from 1 to 11....” shouldn’t the last part be “and 2 Star numbers, from 1 to 11....” after all you don’t know if they’re lucky until the draw takes place? (not used with Lotto, but the terminology seems inconsistent).
MCHandbook 12
Pg3, line 2, “Ireland win...”
Pg3, Q2 same formulation as Q8 in previous handout, I’d make this more precise as well even though the calculations are the same in both cases.
Pg4 table: Straight....unfinished sentence
Pg6 Answer 2 “losing WN” and “losing LS” sounds funny. I think again, here technically you’d try to match possible outcomes for WN-s, LS with your initial picks of numbers? So “losing” should be “not my pick” or “not among the no-s I had picked”. Similarly for “winning” throughout (a)-(f)...
Pg7, 2 Pairs: “which two cards....”
MCHandbook 13
Pg2, 1(b) “..be arraged”
Pg2, Step3, I’d insert “Note that in this way there are 12 stars and bars altogether”
Pg3, (a) shouldn’t it be “more than maximum”?
Pg4, Q3(b) “brining”
Pg5, Q9 (c) “is own”
Pg6, Solution 1(a) “35 way”
Pg 6 last sentence in Answer2
MCHandbook 14
Help for lazy teacher: Here the activity descriptions are addressed to the teacher. If a bit reformulated as addressing directly to the student, the teacher could just print the page and use it as Worksheet?
Pg2, Q1 Pick any number except 0.
Pg6 (4×99∓4×1)
Pg6, Are the kids very confident distributing factors n times? I think this is a bit rushed, “10^=(11−1)^. Each term in this is divisible by 11, other than the last term, which is (−1)^.” I’d go for:
--for an even number of 9-s, 9999…99=11×9×1010…01.
There are as many 1-s in 1010…01 as there are pairs of 9-s in 9999…99.
--for an even number of 0-s, 100…001=11+99…990
99…990 has an even number of 9-s and it’s divisible by 11 as seen in the previous case.
MCHandbook 15
Pg3 “and the difference is: [100(10+−)+90+−−1]−[100(−−1)+90+10+−]” change to sum.
Pg3 last lines and beginning of Pg4: You need only square 2n and 2n+1 which is also consistent to beginning paragraph of Solution 2 ... ok, but I see why you chose 4n, 4n+1 ... in view of next question.
Pg 4 Solution 3 I’d start by suggesting to square the first ... 14 no-s and look for patterns in the last digit, because it’s nice to notice the symmetry by 5 andalso look at
10n-k2=100n2+20nk+k2.
Week 16
Pg1 the paragraph before the last “to first game”
Pg2 Game2 What does “stones” mean here? Should it be coins... Also in solution
Pg2 Game2 This ensures that they will win. “They” is the second player?
Last sentence –is it from the pile with 13 or 14?
Week 17
Last page, last row in Game 6, k should be n?