The University of Melbourne–Department of Mathematics and
Statistics
School Mathematics Competition, 2012
JUNIOR DIVISION
Time allowed: Two hours
These questions are designed to test your ability to analyse a problem and to express yourself
clearly and accurately. The following suggestions are made for your guidance:
(1) Considerable weight will be attached by the examiners to the method of presentation of
a solution. Candidates should state as clearly as they can the reasoning by which they
arrived at their results. In addition, more credit will be given for an elegant than for a
clumsy solution.
(2) The six questions are not of equal length or difficulty. Generally, the later questions
are more difficult than the earlier questions.
(3) It may be necessary to spend considerable time on a problem before any real progress is
made.
(4) You may need to do considerable rough work but you should then write out your final
solution neatly, stating your arguments carefully.
(5) Credit will be given for partial solutions; however a good answer to one question will
normally gain you more credit than sketchy attempts at several questions.
Textbooks, electronic calculators and computers are NOT allowed. Otherwise normal examination conditions apply.
1. The Hunger Games. In the new computer game The Hunger Games, to be released next
month, the French, biology and geography teaching staff at a Melbourne secondary school have
turned into a marauding army of flesh-eating zombies. In order to survive, you must either kill
12 French-teaching zombies, or 8 geography-teaching zombies, or 5 biology-teaching zombies.
Unfortunately, there is no way of telling what subject a particular zombie teaches.
What is the minimum number of zombies that you must kill to be 100% certain you either killed
12 French-teaching zombies, or 8 geography-teaching zombies, or 5 biology-teaching zombies?
Clearly explain your answer.
Solution: The short answer is 12 + 8 + 5 − 3 + 1 = 23. If you kill 22 zombies you might have
the bad luck of having killed 11 French-teaching Zombies, 7 geography teaching zombies and
4 biology-teaching zombies, which is not sufficient to guarantee your survival. But clearly, if
you now kill one more zombie, then this either means you killed at least 12 French-teaching
zombies, or 8 geography teaching zombies or 5 biology-teaching zombies.
2. Woden’s day. The word Wednesday comes from the Old English Wōdensdæg or “Woden’s
day”, after an ancient God believed by some to be the precursor of Father Christmas. The year
2012 is rather special in that the month of February had exactly five Woden’s days.
In what year will this happen next?
Solution: If February 1 falls on a Wednesday, then so do February 8, 15, 22 and 29. Of
course, February 29 only happens in leap years such as 2012. We thus need to find the next
leap year for which February 1 falls on a Wednesday.
Since 365 = 1 + 7 × 52, in a non-leap year the calendar “shifts by one day”, i.e., if it is a Sunday
on February 1 then it will be a Monday on February 1 the year after. Since 366 = 2 + 7 × 52,
in a leap year the calendar “shifts by two days”, i.e., if it is a Sunday on February 1 then it
will be a Tuesday on February 1 the year after. Since there is a leap year every 4 years, in the
next leap year (that is 2016) February 1 will be on a
Wednesday + 2 days + 1 day + 1 day + 1 day = Monday.
Continuing this line of reasoning we get the following table:
Year 2012 2016 2020 2024 2028 2032 2036 2040
Feb 1 Wed Mon Sat Thu Tue Sun Fri Wed
The next year to have 5 Woden’s days in February is 2040.
3. Molly’s mishap. Molly Meldrum suffered a fractured skull while trying to put up Christmas decorations at his home in Highett Street, Richmond late last year. As a result he spent
several months in hospital. When Molly finally returned home in March he could not remember
his house number. Fortunately, he did remember that it was a 3-digit number which is equal
to 11 times the sum of its digits. This was enough information for the taxi driver to deliver
him at exactly the right address.
What is Molly’s house number?
Solution: If Molly lives at xyz Highett Street, Richmond then 100x + 10y + z = 11(x + y + z).
This is also 100x + 10y + z = 11x + 11y + 11z which is the same as 89x = 10z + y. The left-hand
side is a multiple of 89 (i.e., 89 × 1 = 89 or 89 × 2 = 178, etc.) but the largest the largest
number the right-hand side can be is 10 × 9 + 9 = 99. The number x must therefore be 1 so
that we get 89 = 10z + y. This implies that z = 8 and y = 9. Molly thus lives at 198 Highett
Street, Richmond. Note that, indeed, 198 = 11 × (1 + 9 + 8).
3
4. A vampire without bite. Vampire Edward Cullen, played by Hollywood heartthrob
Robert Pattinson, has proven to be especially popular with Melbourne girls. At a recent
screening of Eclipse at Cinema Nova in Carlton, of the people attending the Edward Cullen
swoon-fest, 40% of the males were children, 10% of the children were boys and 80% of the
adults were male.
Given that Cinema Nova can seat up to 100 people, how many girls watched Eclipse that day?
Solution: Let M be the number of men present, W the number of women, B the number of
boys and G the number of girls. Since 40% of the males were children, 60% of the males must
have been men, so that M = (3/2)B. Since 10% of children were boys, 90% of the children
must have been girls, so that B = (1/9)G = G/9. This also implies that M = (3/2)B =
(3/2)(1/9)G = (1/6)G = G/6. Since 80% of the adults were male, 20% of the adults must have
been female, so that W = (1/4)M = (1/4)(1/6)G = (1/24)G = G/24. To summarise
B=
G
,
9
M=
G
,
6
W =
G
.
24
There are now two ways to proceed.
Method 1. The number of girls must be a multiple of 9 = 3 × 3, a multiple of 6 = 2 × 3
and a multiple of 24 = 2 × 2 × 2 × 3. Hence the number of girls must be a multiple of
2 × 2 × 2 × 3 × 3 = 72. Since Cinema Nova only seats 100 there must have been exactly 72 girls
(and 8 boys, 3 women and 12 men, giving a total of 95 people).
Method 2. The total number of people is
G
G
G
95
B+G+M +W = +G+ +
=
G
9
6
24
72
Hence there must have been 72 girls and 95 people in total.
5. That’s puzzling. . . A 7 × 11 jigsaw puzzle has 5 types of pieces:
(a)
(b)
(c)
(d)
(e)
How many pieces of each type make up the puzzle?
Solution: We can first put the edge-pieces of the puzzle. Starting in the bottom-left corner
and moving anti-clockwise we get:
abbbbbbbbcabbbbcabbbbbbbbcabbbbca
where the first and last a denote the same bottom-left piece.
Now note that each b has created a little “blob” pointing towards the middle and each c a
little “hole”. Hence we have 24 blobs and 4 holes, which is 20 more blobs than holes. We now
have to fill up the middle with 77 − 32 = 45 pieces of type d and e. Pieces of type 3 have as
many blobs as they have holes, but pieces of type d have three holes and one blob. Therefore,
if we use 10 pieces of type d and 35 pieces of type e we put in 20 more holes holes than blobs
balancing the 20 more blobs than holes caused by the edge-pieces. To summarise:
Type
(a) (b) (c) (d) (e)
Number of pieces 4 24 4 10 35
Below is an example of a one of the many ways to complete the puzzle.
4
6. Going for gold. Australian hurdler Sally Pearson was named Female Athlete of the Year
in 2011. This year she is hoping to continue her excellent form and win gold at the London
Olympics. But can she think as fast as she can run?
Sally finds herself standing in the middle of the base of an equilateral triangle, with side lengths
100 metres. She must run from her starting position and touch a point on the left-side of the
triangle (excluding the corners), then run to touch a point on the right-side of the triangle
(again excluding the corners), and finally run back to the start. For example, she could run
like
How long is the shortest path Sally can choose, and why?
Solution: The path in the example may also be drawn using reflected triangles as
3
2
2
3
1
Clearly the shortest path is the one corresponding to
3
2
2
1
3
which has length (3/2) times the side length of the triangle. This is (3/2) × 100 = 150 metres.