Test Report: Math for Olympiad from MUMS 3
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These collection of math questions were collected from Melbourne
University Mathematics & Statistics Society.
http://ms.unimelb.au.edu/~mums/ 1. How many ways are there of obtaining 50 cents
change from a large enough supply of 5, 10 and 20 cent coins?
A. 8
B. 10
C. 12
D. 14
E. 24
2. If we add 329 to the three digit number 2x4,
we get 5y3. If 5y3 is divisible by three, what is the greatest
possible value of x?
A. 9
B. 8
C. 15
D. 4
E. 14
3. If t1 = 1 and tn+1
= for all positive
integers n, what is the value of t2004?
4. In ∆ABC, AB = 360, BC
= 507 and CA = 780. M is the midpoint of AC, D is
the point on AC such that BD bisects ÐABC. F is the
point on BC such that BD and DF are perpendicular. The
lines FD and BM meet at E. What is the value of ?
5. In ∆ABC, AB = 5, BC
= 12 and ÐABC = 90o. Arcs of circles are drawn, one with centre A
and radius 5, the other with centre C and radius 12. The two circles
intersect segment AC at M and N respectively. What is the
length of MN?
A. 24
B. 8
C. 16
D. 4
E. 60
6. In how many ways can you colour three edges
of a cube green so that each face has a green edge?
A. 8
B. 6
C. 4
D. 10
E. 12
7. In the lead-up to an election, the
truthfulness of two leading politicians is known to follow the following
pattern: John lies on Monday, Tuesday and Wednesday, and tells the truth on all
other days. Mark, on the other hand, lies on Thursday, Friday and Saturday, and
tells the truth on all other days. One day they both said "Yesterday was
one of my lying days". On which day of the week did they say this?
A. Saturday
B. Friday
C. Monday
D. Wednesday
E. Thursday
8. Let f(n) be the number of
letters used when writing out the digits of the base ten representation of n.
For example, f(27) = 8 since "two seven" has eight letters. What
is the value of f(f(f(… f(216)))) where f
is applied 10 times?
A. 6
B. 4
C. 8
D. 12
E. 14
9. Let t1 = 22. To obtain tn+1,
we square the sum of the digits of tn.
For example, t2 = (2 +
2)2 = 16. Find t2004.
A. 178
B. 98
C. 122
D. 169
E. 128
10. Luke is a fugitive from justice. He steals a
car in Melbourne at 8:00 am and aims to drive it to his freedom. Unfortunately
the Magna he stole is a real bomb and he can only travel at a constant speed of
80 km/h. The police are notified of the theft and commence a chase at 8:30 am
from the location of the theft. They follow the trail of oil which has been
dripping out of the Magna's engine and so follow Luke's route exactly, at a
constant speed of 100 km/h. At what time will Luke be arrested? Remember to
specify in your answer whether the time is am or pm.
A. 10:00 am
B. 11:30 pm
C. 11:30 am
D. 10:30 pm
E. 10:30 am
1.
C
2.
D
To be divisible by 3, the sum of digits
of 5y3 must be divisible by 3, so y = 1, 4 or 7. Now consider the
sum 329 + 2x4. 9 + 4 = 13, so there is a 1 carried over to the tens
digits. 3 + 2 = 5 so the sum of the tens digits and the carried over 1 cannot
be greater than 10. This gives 2 + x + 1 = y, so taking the
maximum value of y = 7, the maximum value of x is 4.
3.
A
By inspection, t1 = 1, t2
= , t3
= , and at this point
one might guess tn = . If not, then calculating the next few terms should make it
very obvious. It is in fact true, so t2004 = is the correct answer.
It can be quite easily proved by induction although it would be a silly thing
to do when only the answer is required.
Unnecessary proof by induction: it is
obviously true for t1 so assume it is true for tn. Then tn+1
= , so tn = is true for all n.
4.
A
Extend FD to meet BA at X.
Applying Menelaus' Theorem to ∆BXF (there is no point providing a
less technical solution as this question is close to impossible without prior
knowledge of this theorem, or at least familiarity with its result), AX
· DF · BC =AB ·DX ·CF, which rearranges to CF
= . DF = DX and BX = BF because BD
bisects ÐXBF and is perpendicular to XF.
Then, .
A bit of algebra yields , and so .
since BD is an
angle bisector. To see this, let ÐABD = ÐCBD = x, ÐADB =
y.
Then and
Divide one equation by the other to get
the required result.
Returning to the problem, since M is the midpoint of AC. Since , solving for DM yields DM = , and . Finally, apply Menelaus to ∆CDF to get DE
· BF ·MC = EF · BC · DM, so
Note that the last question is designed
to be fiendishly hard so that no one can complain
about finishing all the questions and
having nothing to do. It is not expected that anyone
will solve it in the duration of the
competition.
5.
D
By Pythagoras, AC = 13. Also, AM
= 5 and CN = 12, since they lie on circles of that radii at the
respective centres. Hence MN = AM + CN − AC =
4.
6.
A
Each green edge is on two faces, so if
each face of each green edge is unique, then there are 6 faces with green
edges. We thus see that no face can have more than one green edge, or there
would be less than 6 faces with green edges. The problem then becomes
equivalent to the number of ways of finding three pairs of adjacent faces. The
top face
has 4 choices of an adjacent face, and
having selected one, without loss of generality suppose it is the front face.
Then the right face has two choices, the back face or the bottom face, and the
there is only one choice for the last pair. Hence there are 4×2 = 8
possibilities.
7.
E
None of the lying days coincide, so at
least one of them is telling the truth. They cannot
both be telling the truth since that
would mean yesterday is a lying day for both of them.
Thus, exactly one of them is lying, which
means one of them is telling the truth today and
therefore lying yesterday, and the other
is lying today and therefore telling the truth yesterday.
By inspection, the only day which is a
changeover between lying and truthtelling
for both politicians is Thursday.
8.
B
f(216)
= f(65536) = 3 + 4 + 4 + 5 + 3 = 19
f(f(216))
= f(19) = 3 + 4 = 7
f(f(f(216)))
= f(7) = 5
f(f(f(f(216))))
= f(5) = 4
f(f(f(f(f(216)))))
= f(4) = 4
Since f(4) = 4, we can see now
that however many times we apply the function f to the value 4, it will
remain 4. So f(f(f(… f(216) … ))) = 4 if
the function f is applied five or
more times.
9.
D
By inspection, t3 = 49,
t4 = 169, t5 = 256, t6 =
169. Since each term is based entirely on the previous term, any repeated value
means the sequence must cycle. The period is equal to the difference between
the nearest equal terms, here t6 and t4,
and so the period of the cycle is 2. Then, each even subscripted term must be
|169, and each odd subscripted term must be 256, from t4
onwards. Hence t2004 = 169.
10.
E
Suppose that Luke is arrested t
hours after 8:00 am. Then in that time, Luke has travelled 80t
kilometres and the police have travelled 100(t – 0.5) kilometres. Since
these two distances have to be equal, we have the equation:
Therefore, Luke is arrested 2.5 hours
after 8:00am – that is, at 10:30 am.
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