Download RZC-Introduction-Worksheet

Introductory Worksheet
1. [Source: BMO1 – 2012] Isaac places some counters onto the squares of an 8 by 8 chessboard
so that there is at most one counter in each of the 64 squares. Determine, with justification,
the maximum number that he can place without having five or more counters in the same
row, or in the same column, or on either of the two long diagonals.
2. [Source: BMO1] Determine the least natural number 𝑛 for which the following result holds:
No matter how the elements of the set {1, 2, … , 𝑛} are coloured red or blue, there are
integers 𝑥, 𝑦, 𝑧, 𝑤 in the set (not necessarily distinct) of the same colour such that 𝑥 + 𝑦 +
𝑧 = 𝑤.
3. [Source: Cayley] Four types of rectangular tile have sizes 300 mm × 300 mm, 300 mm ×
600mm, 600 mm × 600 mm and 600 mm × 900 mm. Equal numbers of each type of tile are
used, without overlaps, to make a square. What is the smallest square that can be made?
Extra induction questions (if you’ve seen the ‘Epilogue’ slides)
4. Prove that 11𝑛 − 6𝑛 is divisible by 5, ∀𝑛 ∈ ℕ.
5. [Source: Frosty Special] Prove that 6𝑛 + 9𝑛 is divisible by 5 if and only if 𝑛 is odd.
Last updated: 12th September 2014
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Introductory Worksheet - ANSWERS
Note about solutions in the worksheets:
For problems that I’ve obtained from Olympiad or Maths Challenge papers, I tend to work out the
question myself before I look at the model answers (although sometimes these weren’t available, so
I can only hope that what I’ve provided is correct!). The solutions provided are a mix of my own and
the ‘model’ solutions. Sometimes my own solutions uses a different approach from the one
suggested – it’s indicative that there’s often multiple approaches for many questions.
1. Isaac places some counters onto the squares of an 8 by 8 chessboard so that there is at
most one counter in each of the 64 squares. Determine, with justification, the maximum
number that he can place without having five or more counters in the same row, or in the
same column, or on either of the two long diagonals.
The maximum must be at most 32 counters, because we could place 4 counters in each row
at most, and there are 8 columns. All that remains to see if it’s possible to have a valid
arrangement of 32 counters. One possible such arrangement is shown below:
It was slight luck that my ‘intelligent guess’ happened to be the right one. Notice that as the
going across, the pattern repeats every 3 squares horizontally. The consequence is that
because 3 and 8 share no common factors (we’ll see soon that we call them ‘coprime’), the
column the two consecutive vertical counters will be in will be different each time (i.e. first
the 4th column, then the 7th, then the 2nd and so on). This is good because we minimise
column usage, and avoid getting to the 4-counter column limit too soon.
2. Determine the least natural number 𝒏 for which the following result holds:
No matter how the elements of the set {𝟏, 𝟐, … , 𝒏} are coloured red or blue, there are
integers 𝒙, 𝒚, 𝒛, 𝒘 in the set (not necessarily distinct) of the same colour such that 𝒙 + 𝒚 +
𝒛 = 𝒘.
The best way to get your head around this question is to consider simpler cases when 𝑛 is
small. Since also the question says ‘for any colouring’, we want to consider the ‘worst case
scenario’, i.e. try to gradually colour in such a way that we avoid a colouring that where we
have some numbers of the same colour satisfying 𝑥 + 𝑦 + 𝑧 = 𝑤.
For example, suppose 𝑛 = 4. Let the 1 be coloured red say. We’d have to colour 3 blue,
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because if it was red, then we’d have 1 + 1 + 1 = 3, so regardless of what the other
numbers were coloured, we’d have 𝑥, 𝑦, 𝑧, 𝑤 available amongst the reds. We can easily
break the condition by say colouring 1 red, 2 blue, 3 blue and 4 blue. This is because the
condition 𝑥 + 𝑦 + 𝑧 = 𝑤 does not hold with the one number coloured red (1 + 1 + 1 = 3,
but 3 is not a red number), and with the blue numbers, the minimum possible sum of three
numbers is 2 + 2 + 2 = 6, so any sum of three blue numbers will give a number that is not a
blue number.
I tried some higher values of 𝑛 next to see if I could spot a pattern.
Suppose 𝑛 = 8, and 1 was coloured red. 3 would have to be blue. Suppose 2 was also
coloured red. Then we could combine these in sums to get 1 + 1 + 1 = 3, 1 + 1 + 2 = 4,
up to 2 + 2 + 2 = 6. Thus we would have to colour 3, 4, 5 and 6 blue otherwise we’d have a
𝑥, 𝑦, 𝑧, 𝑤 available amongst the red numbers. But at this point, if we coloured 7 and 8 blue,
then there can’t be an blue numbers that satisfy 𝑥 + 𝑦 + 𝑧 = 𝑤, because the minimum sum
with blue is 3 + 3 + 3 = 9, and so we couldn’t have a blue number for 𝑤.
We will encounter a similar problem when 𝑛 = 9 and 𝑛 = 10. If 1 and 2 are coloured red, 38 coloured blue (colouring 3-8 blue avoids being able to have any satisfying 𝑥, 𝑦, 𝑧, 𝑤
amongst the blue, as the minimum sum of three blue numbers is 9, which is not blue) and 9
and 10 red (if there is a 10th), then we know there’s no satisfying 𝑥, 𝑦, 𝑧, 𝑤 for the blue, and
there can’t be any for the red either: if we sum both low and high red numbers, the
minimum possible sum is 1 + 1 + 9 = 11, but 11 is not available since it exceeds 𝑛.
Thus all that remains is to show that for 𝑛 = 11 we can find 𝑥, 𝑦, 𝑧, 𝑤 for any possible
colouring.
As before, without loss of generality1 assume the first number, 1, is red. Now we have two
cases to consider:
a) The second number, 2, is the same colour, i.e. also red. Then as before, we’d have to
colour 3, 4, 5, 6 blue to avoid there being a 𝑥, 𝑦, 𝑧, 𝑤 being available. If we coloured any of 9,
10 or 11 blue, we’d have a 𝑥, 𝑦, 𝑧, 𝑤 available because using the blue 3 to 6, we could get any
sum between 9 and 11. So let’s colour them red. But now we’re guaranteed a 𝑥, 𝑦, 𝑧, 𝑤,
because 1 + 1 + 9 = 11, where all numbers are red. So we’ve managed to show that it’s
impossible to avoid having a colouring where there’s some 𝑥, 𝑦, 𝑧, 𝑤 available for one of the
two colours. Symmetrically, we have the same logic when the first two numbers are blue
instead of red.
b) The second number, 2 is a different colour, i.e. blue. Let’s make 3 blue, because if it was
red than we would have found a 𝑥, 𝑦, 𝑧, 𝑤. This means that if any of 6-9 are blue, we’d have
a 𝑥, 𝑦, 𝑧, 𝑤 because these are all the sums we can get using 2 and 3. So consider them as red.
But then we’re guaranteed a 𝑥, 𝑦, 𝑧, 𝑤 using red numbers, because 1 + 1 + 6 = 8 for
example. So again, we’ve shown that however we colour the numbers, we can’t avoid
having a colouring where there’s some 𝑥, 𝑦, 𝑧, 𝑤 available.
1
‘Without loss of generality’ means we’re considering a specific case, but that doing so doesn’t make our proof
any less general. If we’d instead let the 1 be blue, then we could have used the same logic thereafter but with
the colours switched.
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3. Four types of rectangular tile have sizes 300 mm × 300 mm, 300 mm × 600mm, 600 mm ×
600 mm and 600 mm × 900 mm. Equal numbers of each type of tile are used, without
overlaps, to make a square. What is the smallest square that can be made?
(Official solution) Let us say that one unit is 300 mm, so that the permitted tiles are 1 × 1, 1 ×
2, 2 × 2 and 2 × 3. Since the sides of all the tiles have lengths that are a whole number of
units, any square made out of them will have sides of length 𝑁 that is a whole number of
units. This square has area 𝑁 2 . Also, the total area of one tile of each type is 1 + 2 + 4 +
6 = 13, so is a multiple of 13. The smallest such 𝑁 is 13 itself, so we ask ourselves if such a
tiling is possible with 13 copies of each tile. It can indeed be done, as the example in the
figure shows.
This square, which we have proved to be the smallest possible, measures 13 units on each
side, or 3.9 m × 3.9 m.
[JAF note: Many Year 9 Tiffin students got 6 out of 10 for this question, because although
they proved that the smallest square ‘might be’ 13 of each tile, they didn’t prove that such a
square was possible with this number of tiles)]
4. Prove that 𝟏𝟏𝒏 − 𝟔𝒏 is divisible by 5 ∀𝒏 ∈ ℕ.
Base Case: Show it’s true for 𝑛 = 1: 11 − 6 = 5, so true.
Inductive Step: Assume it’s true for 𝑛 = 𝑘. Suppose now 𝑛 = 𝑘 + 1.
11𝑘+1 − 6𝑘+1
= 11 ⋅ 11𝑘 − 6 ⋅ 6𝑘
= 6(11𝑘 − 6𝑘 ) + 5 ⋅ 11𝑘
The first term in the sum is divisible by 5 as we assumed 11𝑘 − 6𝑘 was. And the second term
is clearly divisible by 5.
5. Prove that 𝟔𝒏 + 𝟗𝒏 is divisible by 5 if and only if 𝒏 is odd.
This question is all about being rigorous with your proof by ensuring it’s ‘complete’.
Note that we can deal with the ‘if and only if’ by showing that 6𝑛 + 9𝑛 is divisible by 5 if 𝑛 is
odd, and also (very importantly!) not if 𝑛 is even. This is because “if is 𝑛 is even, 6𝑛 + 9𝑛 is
not divisible by 5” is equivalent to “if 6𝑛 + 9𝑛 is divisible by 5, 𝑛 is odd”. We’ll see more of
negating/reversing conditional statements like this in the Logic module.
Proof Part 1: If 𝒏 is odd, 𝟔𝒏 + 𝟗𝒏 is divisible by 5.
If 𝑛 is odd, represent it as 𝑛 = 2𝑘 + 1. Our problem becomes equivalent to showing
62𝑘+1 + 92𝑘+1 is divisible by 5 for all positive integers 𝑘.
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Base Case: When 𝑘 = 1, we have 216 + 729 = 945 which is divisible by 5.
Inductive Step: Assume true for 𝑛 = 𝑘. Now show true for 𝑛 = 𝑘 + 1.
62(𝑘+1)+1 + 92(𝑘+1)+1
= 62𝑘+3 + 92𝑘+3
= 36 ⋅ 62𝑘+1 + 81 ⋅ 92𝑘+1
= 36(62𝑘+1 + 92𝑘+1 ) + 45 ⋅ 92𝑘+1
The first term is divisible by 5 because we assumed 62𝑘+1 + 92𝑘+1 was. And the second one
is because 45 is.
Proof Part 2: If 𝒏 is even, 𝟔𝒏 + 𝟗𝒏 is not divisible by 5.
Note that a counterexample is not sufficient: we need to show that for any even number,
the statement does not hold.
If 𝑛 is even, then let 𝑛 = 2𝑘. Thus we need to prove that 62𝑘 + 92𝑘 = 36𝑘 + 81𝑘 is never
divisible by 5 for any integer 𝑘.
Base Case: When 𝑘 = 1, 36 + 81 = 117 which is not divisible by 5.
Inductive Step: Assume it’s not divisible for 𝑘, now show it’s still not divisible for 𝑘 + 1.
36𝑘+1 + 81𝑘+1
= 36 ⋅ 36𝑘 + 81 ⋅ 81𝑘
= 36(36𝑘 + 81𝑘 ) + 45(81𝑘 )
Now, the first term is not divisible by 5 because neither 36 nor 36𝑘 + 81𝑘 is. The second
term clearly is divisible by 5 because 45 is. And clearly a number not divisible by 5 plus a
number divisible by 5 will give a number not divisible by 5.
You’ll see this problem again in one of the Number Theory module worksheets. The
complete proof can be reduced to just two lines using modular arithmetic!
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