Solutions to Maths Challenge #1

(1)

A palindrome is a word, phrase or number that reads the same backwards and forwards, eg “Was it a rat I saw?”, and 59695.

The last palindromic years were 1991 and 2002 (a gap of 11 years).

Since the year, 1AD what has been the largest gap between palindromic years?

Since 1 AD, the maximum gap between palindromic years is 110 years, and it has

happened 11 times – for example between 1881 and 1991.

Multiplying the original equation by x gives: x

3

+ x

2

+ x = 0

⇒

x

3

– 1 = 0

⇒

x

3

= 1

But the original equation, when re-arranged gives: x

2

+ x = -1

Solutions to Maths Challenge #1

For Year 7 to 10

11212Y12:

For Year 11 to Y13:

Given that x

2

+ x + 1 = 0, find x

3

(2)

Solution to Maths Challenge #2

Find the sum of all the proper fractions with a denominator of 100 or less. Include

fractions such as 2/4, 26/60 etc which aren't in their simplest form.

Look at the fractions you are adding:

1/2 1/3 + 2/3 1/4 + 2/4 + 3/4 1/5 + 2/5 + 3/5 + 4/5 1/6 + 2/6 + 3/6 + 4/6 + 5/6 ….

= 0.5 = 1 = 1.5 = 2 = 2.5

The partial sums are increasing by 1/2 each time. The last one will be:

1/100 + 2/100 + 3/100 + ……. + 98/100 + 99/100

and this will sum to

49.5

(3)

Solution To Maths Challenge 3

A Logic Problem

This is a conversation I overheard recently:

X: You know, I don’t know how old your three kids are. Y: If you multiply their ages, you get 36.

X: I still don’t know how old they are.

Y: The sum of their ages is your house number. X: I still don’t know how old they are!

Y: Well ... the eldest child has red hair. X: Ah, now I know how old they are!

What logic has X used, and how old are the three children?

Here are the only sets of three numbers whose product is 36. The sum of each triple is also given:

Numbers Sum 1 1 36 38 1 2 18 21 1 3 12 16 1 4 9 14 1 6 6 13 2 2 9 13 2 3 6 11 3 3 4 10

Y says the sum of the ages adds up to X’s house number, yet X says he still doesn’t know their ages. We can certainly assume that X knows his own house number - so this means the ages must be one of the triples in highlighted bold type above (they both sum to the same number – that’s why X didn’t have enough information to go on). Y also mentions that there is an eldestchild. That rules out the possibility of 1, 6, 6.

This means the three children were aged

2, 2

and

9

Correct Solutions from:

(4)

The minimum number of queens you can place on a chessboard so that every square is either

occupied or under threat is

5 . One possible arrangement is shown below:

Solution to Maths Challenge #4

Every square is either occupied by a queen or threatened by one.

In the arrangement shown, no queen

threatens another. Interestingly there is an arrangement where every queen is under threat from another queen. Can you find this?

(5)

Solution To Maths Challenge #5

This is easier than it looks! We may not know any of the horizontal

lines in the top half of the diagram, but they do all add up to 21cm.

Likewise the sum of all the vertical lines in the left-hand side of the

diagram add up to 25cm.

The perimeter is 2 x (21cm + 25cm) =

92cm

Correct Answers from:

Calista (6K, Sapi) and Debbie (6S, Gaya)

Find x if: 3 =

x

₊

x

₊

x

₊

x

₊

...

Squaring both sides gives:

9 =

X

+

x

₊

x

₊

x

₊

x

₊

...

which simplifies to:

9 = X + 3

⇒

X = 6

Correct Answers from:

Evan (9S, Sapi)

For Years 7 to 9

For Years 10 to 13

Calculate the Perimeter of this shape (all angles are right-angles).

(6)

Solution To Maths Challenge #6

Two men and two boys wish to cross a river. Their small canoe can carry the weight of only one man or two boys.

Find the minimum number of times the canoe must cross the river to get all four people to the opposite shore.

Here are the movements:

Move First Shore Canoe Opposite Shore

There are

other ways to

do this, but

you won’t do it

in less than

9 moves!

- M M B B 1 M M (B B) → 2 M M ← B B 3 M B M → B 4 M B ← B M 5 M (B B) → M 6 M ← B M B 7 B M → M B 8 B ← B M M 9 (B B) → MM - B B M M

Correct Answers from:

Calista (6K, Sapi), Debbie (6S, Gaya), Max (5S, Gaya) Amiel (5K, Sapi), Brandon (5K, Gaya)

The number 296 – 1 has two factors between 60 and 70. What are they?

2

96

– 1 = (2

48

– 1)(2

48

+ 1)

= (2

24

– 1)(2

24

+ 1)(2

48

+ 1)

= (2

12

– 1)(2

12

+ 1)(2

24

+ 1)(2

48

+ 1)

= (2

6

– 1)(2

6

+ 1)(2

12

+ 1)(2

24

+ 1)(2

48

+ 1)

= 63 x 65 x (2

12

+ 1) x (2

24

+ 1) x (2

48

+ 1)

For Years 5 to 9

For Years 10 to 13

So the answer is:

(7)

Solution To Maths Challenge #7

A mixture of 30 litres of paint is 25% red tint, 30% yellow tint and

45% water. Five litres of yellow tint are added to the original mixture.

What is the percentage of yellow tint in the new mixture now?

The original mixture must contain 30% of 30 litres as yellow tint.

This is 0.3 x 30 = 9 litres

If we now add 5 more litres of yellow tint, we’ll have 14 litres in a total

volume of 35 litres

Thus the new percentage is

!"_!"

x

100 %

or

!"

%

Correct Solution from: So Hee (7S, Sapi), Calista (6K, Sapi), Debbie (6S, Gaya)

A 3 4 B 5 C

Let the radius of the smallest circle be

r1, the radius of the middle circle be r2 and the radius of the biggest circle be r3

This gives:

r1 + r2 = 3 r1 = 1

r1 + r3 = 4 r2 = 2 ⇒ Total Area = π(1)2 + π(2)2 + π(3)2 =

14π

r2 + r3 = 5 r3 = 3

Correct Solution from: Ethan (10S, Gaya)

For Years 5 to 9

For Years 10 to 13

The 3 circles shown just touch each other.

A, B and C are the centres of the circles.

A 3-4-5 triangle is drawn with vertices at A, B and C

(8)

Solution To Maths Challenge #8

1. There is more than one alien.

2. Each alien has the same number of heads. 3. Each alien has more than one head.

4. The total number of heads in the room is between 200 and 300.

5. If you knew the total number of heads in the room you would know how many aliens there are.

How many aliens are there?

The really important clue is number 5

This clue would rule out a number like 240 for the number of heads, for example. Because if there were 240 heads in the room, it could be 20 aliens with 12 heads each, or 80 aliens with 3 heads each or 10 aliens with 24 heads each etc.

In other words, the problem is really one about factors. We can’t have a highly composite number like 240 for the number of heads in the room. Also, we can rule out a prime number for the number of heads because we’re told that there is more than one alien, and that each alien has more than one head.

So, what are we left with? We are left with the squareof a prime number! And the only one in the interval given is 289.

So there are 289 heads in the room, which means that there are 17 aliens each with 17 heads.

For All Years

In a room there are a number of aliens from another planet.

The following are true:

(9)

Split the chessboard up into 4 congruent (i.e. same shape,

same size) pieces, each of which contains one

symbol.

Correct Solution from:

Miche (Y12, Gaya)

Solution to Maths Challenge #9

For All Years

(10)

Solution To Maths Challenge #10

You have a simple digital clock, which gives the time using 24-hour

clock notation (hours and minutes only, no seconds).

Between 0000 and 0100 you will see a 5 on the display for 15 minutes (one minute for each of: 0005, 0015, 0025, 0035, 0045 and 10 minutes between 0050 and 0059). This is also true for nearly every hour in the day.

The exceptions being that you will see a 5 for the whole hour between 0500 and 0559 and the whole hour between 1500 and 1559. Thus the total time is:

22 x 15 minutes and 2 x 1 hour, giving a total of 7½ hours

Correct Solution from: Brandon (5K, Gaya) and others (who didn’t write their name!

Find the next two numbers in the sequence:

2, 6, 30, 210, 2310 …

The key to solving this problem is to write the numbers as products of their prime factors. i.e. the numbers are:

2, (2 x 3), (2 x 3 x 5), (2 x 3 x 5 x 7), (2 x 3 x 5 x 7 x 11) ….. So the next 2 numbers are:

(2 x 3 x 5 x 7 x 11 x 13) and (2 x 3 x 5 x 7 x 11 x 13 x 17)

In other words … 30 030 and 510 510

Correct Solution from: Tina (10S, Sulug) and Evan (9S, Sapi)

For Years 5 to 9

For Years 10 to 13

For how much time in the

day will you be able to see at

least one 5 in the display?

(11)

Solution To Maths Challenge #11

Which of these shapes is the odd one out:

A B C D E

At first sight, more than one answer seems possible.

You could say the answer is B (the only shape without a thick black border). You could say the answer is C (the only shape that’s not a square).

You could say the answer is D (the only shape that’s green in the middle). You could say the answer is E (a significantly smaller shape).

The only shape that doesn’t seem to be the odd one out is A.

This makes A the odd one out!

Correct Solution from: Sandro (11K, Gaya)

Find the next number in the sequence:

0, 1, 2, 4, 5, 6, 8, 10, 40, 46, 60, 61, 64, 80, 84,?

Write these numbers out: ZERO, ONE, TWO, FOUR, FIVE, SIX etc What about the missing numbers – THREE, SEVEN, NINE etc?

You soon see that the numbers given never use the same letter twice! The next number you can say that about is the number 5000 (FIVE THOUSAND)

For All Years (A)

For All Years (B)

You must give a reason for your answer!

(12)

Solution To Maths Challenge #12

Find three 2-digit prime numbers so that:

The mean (average) of any two of them is prime.

And

…

… The mean (average) of all three of them is prime.

The numbers are:

11, 47

and

71 Check: the mean of all three is 43 (a prime), and the mean of any pair

is also prime.

34kg of fresh grapes contains 0.80 x 34kg = 27.2kg water.

This means there will be (34 – 27.2)kg = 6.8kg of “grape essence”, or

waterless content.

This 6.8kg of grape essence can be used to make dried grapes.

The dried grapes would weigh 6.8kg (+ 15% more for water content).

That is, we would get 6.8 x 1.15kg of dried grapes.

So the answer is 8kg of dried grapes

Correct Solutions from:

Sandro (11K, Gaya), Sam (Y12, Gaya), Stuart (Y12, Sulug)

For Years 5 to 9

For Years 10 to 13

Fresh grapes contain 80 percent water by weight,

whereas dried grapes contain 15 percent water by

weight.

How many kg of dried grapes can be obtained

from 34kg of fresh grapes?

(13)

Solution To Maths Challenge #13

3 friends are captured by the evil Kurt. He puts each friend in a cell of his own. He also puts some apples into each cell. Each friend can count the number of apples in his own cell, but not his friends’. It is known that:

• The number of apples in each cell is different

• Each cell has at least 1 apple

• No cell has more than 9 apples

To escape, one of the 3 friends has to tell Kurt exactly how many apples there are in total in the three cells. Each friend is allowed to ask one question, which Kurt will answer truthfully “yes” or “no”. Everyone hears the questions and the answers. Here’s how the conversation has gone so far:

Friend A: “Is the total number of apples an even number?”

Kurt: “No”

Friend B: “Is the total number of apples a prime number?”

Kurt: “No”

It’s now friend C’s turn ... What question should C ask so the friends can escape the cells?

First, a bit of arithmetic … the total number of apples must be somewhere between 6 and 24 (can you see why?)

The fact that the total isn’t even and isn’t prime means the total must be 9, 15 or 21 (can you see why?). And all three friends will know this!

Now for the answer, and the logic behind it … The question C should ask is “Is the total 15?”

If Kurt answers “yes”, then clearly the answer is 15 (Kurt never lies).

If Kurt answers “no” then the answer is 9 or 21. But look at what this means … For there to be a total of 9 apples, there would have to be these amounts in the three cells: (1, 2, 6) or (1, 3, 5) or (2, 3, 4)

For there to be a total of 21 apples, there would have to be these amounts in the three cells: (5, 7, 9) or (6, 7, 8).

So if anyone has 7 or 8 or apples in his cell, he knows the total is 21 and will say so. And if anyone has 1 or 2 or 3 or 4 apples in his cell, he will know the total is 9 and say so!

Correct Solution from: Sandro (11K, Gaya)

(14)

Solutions to Maths Challenge #14

Here is a simple and famous puzzle. All you have to do is find out what number the question mark is standing for.

At first sight you are led to think that there is a simple rule – the

number in the circle being pointed to is the difference of the 2

numbers above.

But this doesn’t work all the time!

So there must be another rule at work (one that

does

work all the

time).

And there is – a very simple one, too.

The number being pointed to is the sum of all the digits of the

numbers pointing to it!

So the question mark is standing for

12.

Correct Solutions from: Ethan (10S, Gaya) For All Years

Be careful! The 7 at the bottom of the diagram is NOT a mistake!

(15)

Solutions to Maths Challenge #15

The Problem of the Year (2018)

Using the digits 2, 0, 1 and 8 exactly once, and any mathematical symbols

you know, generate as many of the numbers from 1 to 40 as you can (you might not get them all)!

Extra credit is earned if you use the digits in the order given.

For example:

11 = 20 – 8 – 1 scores 1 point, whereas 11 = 20 – 1 – 8 scores 2 points

Target Arithmetic Score Target Arithmetic Score

1 2 x 0 + 18 2 21 2 + 0! + 18 2 2 20 – 18 2 22 21 + 80 1 3 2 + 0 + 18 2 23 20 + 1+8 2 4 2 + 0! + 18 2 24 (2 + 0 + 1) x 8 2 5 2 + 0 + 1+8 2 25 (2 + 1) x 8 + 0! 1 6 2 + 0! + 1+8 2 26 28 – 0! – 1 1 7 2 + 0 – 1 + 8 2 27 20 – 1 + 8 2 8 2 x 0 + 1 x 8 2 28 20 + 1 x 8 2 9 2 x 0 + 1 + 8 2 29 20 + 1 + 8 2 10 2 + 0 + 1 x 8 2 30 28 + 0! + 1 1 11 20 – 1 – 8 2 31 12 20 x 1 – 8 2 32 82_{÷ (0! + 1)} ₁ 13 20 + 1 – 8 2 33 14 (2 + 0 + 1)! + 8 2 34 2 x (18 – 0!) 1 15 (2 + 0!)! + 1 + 8 2 35 2 x 18 – 0! 1 16 (2 + 0) x 1 x 8 2 36 2 x 18 + 0 1 17 20 – 1+8 2 37 2 x 18 + 0! 1 18 2 x 0 + 18 2 38 20 + 18 2 19 20_{+ 18} ₂ ₃₉ _{80 ÷ 2 – 1} ₁ 20 2 + 0 + 18 2 40 80 ÷ (2 x 1) 1

This would give a score of 65/80. Can you do better? For All Years

(16)

Solutions to Maths Challenge #16

1. The number of zeroes is

425 .

2. The last digit of 2018

2018

is a

4.

3. 0.201820182018 … = 2018

9999

4.

Number(s) How many of these there are

2 1

20 – 29 10

200 – 299 100

2 000 – 2 999 1 000

20 000 – 29 999 10 000

Best Solutions from: Ethan (10S, Gaya), Stuart (Y12, Sulug), Sam (Y12, Gaya)

So by the time the number 2 999 is written down, this is

the 1,111th_{number beginning with 2. The 2018}th_number

is thus the 907th_{number in the 20 000 – 29 999 range}

which is 20 906.

Each “decade” that contains numbers ending in 2 and 5 produces a zero, each multiple of 10 produces a zero, each multiple of 100 produces 2 and each multiple of 1000 produces 3.

There is a very simple pattern – the last digit of 2018n follows the

(17)

One More Problem of the Year (2018)

Using each of the digits from 1 to 9 exactly once, and

any mathematical symbols you know, create the

number 2018

(special credit if the digits are in strict 1

to 9 order

).

There are many possible solutions. My favourite is:

1

2 + 3 – 4

×

5 + 678

×

9 ... and this of course uses the digits in order.

Good solution from:

Sam (Y12, Gaya) and Stuart (Y12, Sulug)

Solution to Maths Challenge #17

(18)

Solutions to Maths Challenge #18

Final Problem of the Year 2018

Here is a simple rule:

If a number is even, divide it by 2

If a number is odd, multiply it by 3 and add 1

E.g. starting with 12, the first few terms we get are: 12, 6, 3, 10, 5, 16, …

Applying the rule(s) we get the following number chain:

2018 ! 1009 ! 3028 ! 1514 ! 757 ! 2272 ! 1136 ! 568 ! 284 ! 142 ! 71 !214 ! 107 ! 322 ! 161 ! 484 ! 242 ! 121 ! 364 ! 182 ! 91 ! 274 ! 137 ! 412 ! 206 ! 103 ! 310 ! 155 ! 466 ! 233 ! 700 ! 350 ! 175 ! 526 ! 263 ! 790 ! 395 ! 1186 ! 593 ! 1780 ! 890 ! 445 ! 1336 ! 668 ! 334 ! 167 ! 502 ! 251 ! 754 ! 377 ! 1132 ! 566 ! 283 ! 850 ! 425 ! 1276 ! 638 ! 319 ! 958 ! 479 ! 1438 ! 719 ! 2158 ! 1079 ! 3238 ! 1619 ! 4858 ! 2429 ! 7288 ! 3644 ! 1822 ! 911 ! 2734 ! 1367 ! 4102 ! 2051 ! 6154 ! 3077 ! 9232 ! 4616 ! 2308 ! 1154 ! 577 ! 1732 ! 866 ! 433 ! 1300 ! 650 ! 325 ! 976 ! 488 ! 244 ! 122 ! 61 ! 184 ! 92 ! 46 ! 23 ! 70 ! 35 ! 106 ! 53 ! 160 ! 80 ! 40 ! 20 ! 10 ! 5 ! 16 ! 8 ! 4 ! 2 ! 1

Then 4 ! 2 ! 1 ! 4 ! 2 ! 1 etc. etc.

The first 1 is the 113th number in the chain. The 114th number is 4, the 115th is 2, the 116th is 1 again etc.

Thus, the 2018

th

number will be a 1

Starting with 2018, what is the

(19)

Solution To Maths Challenge #19

If …

Germany = 73

Tonga = 47

Fiji = 24

Azerbaijan = 77

If we assign the simple cipher:

A=1, B=2, C=3, D=4 …….. Y=25, Z=26 then add the numbers in the countries’ names we get: Germany = 7 + 5 + 18 + 13 + 1 + 14 + 25 (which is 83) Tonga = 20 + 15 + 14 + 7 + 1 (which is 57) Fiji = 6 + 9 + 10 + 9 (which is 34) Azerbaijan = 1 + 26 + 5 + 18 + 2 + 1 + 9 + 10 + 1 + 14 (which is 87)

So, Malaysia must be (13 + 1 + 12 + 1 + 25 + 19 + 9 + 1) – 10 =

71

Correct Solution from:

How many factors does 60

5

have?

605 can be written as 2103555

In other words, 605 can be divided by 20 through 210, by 30 through 35 and by

50 through 55. This gives us 11 x 6 x 6 or 396 factors

Correct Solution from: Ethan (10S, Gaya) and Eric (10K, Sapi)

For Years 5 to 9

For Years 10 to 13

Break the code and find out

what number Malaysia is!

And these

numbers are all 10 less than the

numbers in the code ….

(20)

Solutions to Maths Challenge #20

In the alphametic below, each letter stands for one and only one digit. Can you find out which letter

stands for which digit?

!"##$%&

= KISS

Logic tells us, for example, that S cannot stand for the digits 0, 1, 5 or 6. Can you see why?

Also, N couldn’t possibly be 2, 3, 7 or 8. Why?

In addition, KISS < 3162 (If KISS was bigger, then squaring it would result in an 8-digit number)

From this point on, all we need is to try squaring numbers with a calculator, bearing in mind the

above restrictions. Soon we hit upon the answer:

4133089

= 2033

Correct answer from: Dania (11S, Gaya) and Sandro (11K, Gaya) For All Years

(21)

Solution to Maths Challenge #21

Bananas Today, Bananas Tomorrow!

There’s a certain amount of trial and error, but it soon becomes obvious when you make a mistake!

Here is the correct order for the fractions:

!!_!"

,

!_!

,

!_!

,

!_!

,

!_!

,

!_!

Check that this means he ended up with one banana!

Correct Solution from:

John (9K, Gaya)

A monkey has 75 bananas. Each day, he kept a fraction of his

bananas, gave the rest away and ate one.

These are the fractions he decided to keep:

1

2 ,

1

4 ,

3

4 ,

3

5 ,

5

6 ,

11

15 In what order did he keep these fractions, given that he ended

with one banana?

(22)

Solution To Maths Challenge #22

Turn the 5-minute and 11-minute hourglasses upside down at the same time. When the 5-minute one runs out, that’s when we start the 13-minute interval. 6 minutes later, the 11-minute hourglass runs out.

Immediately this happens, the 7-minute one is started.

When the 7-minute hourglass runs down, 13 minutes have elapsed since the start.

Since 1864 = 43.147, de Morgan must have been 43 or less in the year when the statement was true.

412 = 1681. If this were the age that he mentioned, he would have been born in 1640 and would have been 224 years old in 1864. Impossible

422_{= 1764. If this were the age that he mentioned, he would have been born in 1722}

and would have been 142 years old in 1864.

432 = 1849. If this were the age that he mentioned, he would have been born in 1806

and would have been 43 years old in 1849 – just a few years before he made the statement. This is, realistically, the only feasible answer.

For Years 5 to 8

How can you measure 13 minutes

exactly

with a

5-minute, a 7-5-minute, and an 11-minute hourglass?

Augustus de Morgan wrote in 1864:

"

At some point in my life, the square of my age was

the same as the year.

"

(23)

Solution to Maths Challenge #23

Let

N

represent 4

9

+ 6

10

+ 3

20

. Then

N

= 2

18

+ (2

10

)(3

10

) + 3

20

= (2

9

)

2

+ 2(2

9

)(3

10

)+ (3

10

)

2

= (2

9

+ 3

10

)

2

i.e. N is a square number, therefore it cannot be prime.

For Years 5 to 8

For Years 9 to 13

Prove (without using a calculator) that the number 4

9

+ 6

10

+ 3

20

is

not

prime.

The combination is:

(24)

Solution to Maths Challenge #24

Find the difference between the sum of the first 1,000,000 positive even

numbers and the sum of the first 1,000,000 positive odd numbers.

There is a very easy pattern than can help us here, and here is a brief snapshot of that pattern:

Sum of first 5 even numbers = 2 + 4 + 6 + 8 + 10 = 30 Difference = 5

Sum of first 5 odd numbers = 1 + 3 + 5 + 7 + 9 = 25

Sum of first 10 even numbers = 30 + 12 + 14 + 16 + 18 + 20 = 110 Difference = 10

Sum of first 10 odd numbers = 25 + 11 + 13 + 15 + 17 + 19 = 100

This leads to the (correct) assumption that the difference between the first n even numbers and

the first n odd numbers is n.

So the difference between the first 1 000 000 even numbers and the first 1 000 000 odd numbers

is 1 000 000.

Correct Answers from: All Grades

(25)

Solution to Maths Challenge #25

There were 4 football matches taking place on Sunday afternoon.

John thought the winners would be Liverpool, Spurs, Chelsea and Everton.

Sue thought that Everton, Man City, Chelsea and Southampton would be the winning teams.

Nick said Man Utd, Man City, Spurs and Everton would all win their matches and that Arsenal would

score no goals.

Who played whom?

There are 8 teams altogether.

Everton can’t possibly play Liverpool, Spurs, Chelsea, Man City, Southampton or Man Utd.

This means they must play Arsenal.

Chelsea can’t (obviously) play Everton or Arsenal. Nor can they play Spurs, Liverpool,

Southampton or Man City.

This means they play Man Utd.

Man City can’t be playing Spurs, Chelsea, Everton, Arsenal, Southampton or Man Utd.

This means they play Liverpool.

And that leaves

Spurs playing Southampton

.

Correct Answers from:

Sandro (11K, Gaya), Miche (Y12, Gaya), Ethan (10S, Gaya)

(26)

Solution To Maths Challenge #26

How many five-digit numbers greater than 70,000 are mountain numbers?

As we are looking for mountain numbers greater than 70,000, the first three digits must be 7, 8, and 9 in that order.

So the numbers are of the form 789ab, where a > b and a < 9.

We start with the greatest possible value for a, which is 8, and work down to the

least possible value, which is 1.

For a = 8, the value of b can be any digit from 7 through 0, which is 8 choices.

⋅ ⋅ ⋅

For a = 1, the value of b can only be 0, which is one choice.

A total of 8 + 7 + 6 + ... + 1 + 0 = 36 possible numbers fit the criteria.

A five-digit positive integer is a "mountain number" if

the first three digits are in ascending order and the last three digits are in descending order.

For example, 35,761 is a mountain number, but

(27)

Solution To Maths Challenge #27

When all the numbers from 1 to 1000 are written out, which digit (if any) is used least often?

The answer is the

0 . By way of explanation, look for example at the 90 numbers from 10 to 99.

Although there is an equal distribution of digits in the 2

nd

position, no zeroes ever occupy the

first position. A similar thing happens with the 900 numbers between 100 and 999.

A gambler has 2 coins in his pocket—one fair coin and one two-headed coin. He selects a coin at random and flips it twice. If he gets two heads, what is the probability that it was the fair coin he had selected? The quickest way to answer this is to consider this: The fair coin gives 2 Heads (in 2 throws) ¼ of the time.

The trick coin gives 2 Heads (in 2 throws) all the time. So if you got 2 Heads, it’s 4 times more likely to

have come from the trick coin than the fair coin.

In other words, the required probability is

1/5

Correct Solution from: John (Y9, Gaya)

For Years 8 and below

(28)

(29)

Solution To Maths Challenge #28

There are 30 students in a class and they all line up. You find that the largest number of consecutive boys in a row is 4.

What is the maximum number of boys that could be in the class?

The key here is to appreciate the use of the word “maximum”.

The answer is 24.

The six girls in the class are then interspersed among the boys to

create 6 groups of 4. If there were more than 24 boys (and so less

than 6 girls) at least one group of boys would be greater than 4 in

number.

Here’s an arrangement with 25 boys and 5 girls:

BBBBGBBBBGBBBBGBBBBGBBBBGBBBBB

You can see that there is a group of 5 boys at then end.

How many factors does 60

5

have?

60

5

can be written as 2

10

3

5

In other words, 60

5

can be divided by 2

0

through 2

10

(

11 different

numbers), by 3

0

through 3

5

(

6 different numbers) and by 5

0

through

5

(also

6 different numbers).

This gives us 11 x 6 x 6 or

396 factors.

For Years 8 and below

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Solutions to Maths Challenge #29

Which is the better fit – a square peg in a round hole, or a round peg in a square hole?

A round peg in a square hole: A square peg in a round hole:

So a round peg in a square hole is a better fit.

Correct Solutions from: John (9K, Gaya) and Ethan (10S, Gaya)

The percentage of space “wasted” is [(2a)2_–

πa2] x 100%

(2a)2

_{= 4 –}

π x 100%

4 = 21.46% approximately

The radius of the circle must be √2!

The percentage of space “wasted” is [2πa2 - (2a)2] x 100%

2πa2 _{= 2}

π − 4 x 100%

2π = 36.34% approximately The shaded area is

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Solutions to Maths Challenge #30

The key here is to notice that, starting from 1, the leading diagonal (going down and right) contains the odd squares.

Similarly, starting from 4, the leading diagonal going up and left contains the even squares.

These are highlighted in the diagram to the left. Now 442_{= 1936 and 45}2_{= 2025}

This gives us a clue as to where 2018 will be in the expanded diagram … 31 17 16 15 14 13 30 18 5 4 3 12 29 19 6 1 2 11 28 20 7 8 9 10 27 21 22 23 24 25 26 37 36 35 34 33 32 31 38 17 16 15 14 13 30 39 18 5 4 3 12 29 40 19 6 1 2 11 28 41 20 7 8 9 10 27 42 21 22 23 24 25 26 51 43 44 45 46 47 48 49 50

All the positive integers are written in the cells of a square grid, as shown. Starting from 1, the numbers spiral

anticlockwise.

The first part of the spiral is shown in the diagram.

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1937 1936 1937 1938 . . . . 37 36 35 34 33 32 31 38 17 16 15 14 13 30 39 18 5 4 3 12 29 40 19 6 1 2 11 28 . _. 41 20 7 8 9 10 27 . _. 42 21 22 23 24 25 26 51 43 44 45 46 47 48 49 50 2018 …. … 2023 2024 2025 2201 … … 2206 2207 2208 2209

Thus the number beneath 2014 is

2201

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Solution to Maths Challenge #31

The diagram shows an equilateral triangle with its corners at

the midpoints of the sides of a regular hexagon.

The trick is to draw some extra lines like in the diagram below:

For All Years

What fraction of the

hexagon is shaded?

It is clear from this diagram that 9 of the 24 small

equilateral triangles are shaded.

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Solution to Maths Challenge #32

4 black cows and 3 brown cows give as much milk in five days as 3

black cows and 5 brown cows give in four days.

Which cow gives more milk – the black or the brown?

Assume the black cow gives x litres/day, and the brown gives y litres/day.

Thus, 5(4x + 3y) = 4(3x + 5y)

⇒

20x + 15y = 12x + 20y

⇒

8x = 5y

⇒

y > x i.e The brown cow gives more milk

Four men, one of whom was known to have committed a certain

crime, made the following statements when questioned by the police:

Archie:

“Dave did it”.

Dave:

“Tony did it”.

Gus:

“I didn't do it”.

Tony:

“Dave lied when he said I did it”.

If only one of these four statements is true, who was the guilty man?

If the culprit

was …

Archie’s

Then these men’s statements would be …

Dave’s

Gus’s

Tony’s

Archie

False

True

Dave

True

False

True

Gus

False

True

Tony

False

True

False

The only combination in which three statements are false and one

true happened when

Gus committed the crime.

For Years 8 and below

For Years 9 and above

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Solutions to Maths Challenge #33

The Borneo Cup

This is a trophy contested by 4 teams. Here is the league table after all the teams have played each other once:

Team Goals Scored Goals Against Points

Kudat 7 1 9

Tuaran 3 3 4

Ranau 2 3 4

Sandakan 1 6 0

Some obvious conclusions from the table:

• Kudat won all their games, and Sandakan lost all their games

• Ranau (having conceded 3 goals against Kudat) never conceded another

goal so must have won a game and drew a game, in order to have 4 points. Given Ranau only scored 2 goals, their other results must have been a 0-0 draw and a 2-0 win. Ranau’s win was against Sandakan (and the draw against Tuaran) so Sandakan’s goal was not against Ranau

• Tuaran won 1, drew 1 and lost 1 game. The win was against Sandakan,

and the draw was against Ranau – which must have been 0-0

Thus we now know all of the results, in terms of wins/losses and draws, and we know all three results involving Ranau. It doesn’t take too much

experimentation to end with:

Kudat 3 – 0 Ranau Tuaran 2 – 1 Sandakan Kudat 2 – 0 Sandakan Tuaran 0 – 0 Ranau Kudat 2 – 1 Tuaran Sandakan 0 – 2 Ranau

Correct Solutions from:

If I tell you that Kudat beat Ranau 3-0, can you work out the score in every other match?