Western Australian Junior Mathematics Olympiad 2011: Individual Questions
Western Australian Junior Mathematics Olympiad 2011: Individual Questions
Western Australian Junior Mathematics Olympiad 2011: Individual Questions
100 minutes
1
x
2
+ 1 y 6?
[3 marks]
P
6m
1m B
[3 marks]
8. Two friends are training at the same circular track, running in opposite directions. Chris takes 112 seconds to run a complete lap,
but finds that he is meeting Sophie every 48 seconds.
How many seconds is Sophie taking for each lap?
[3 marks]
9. On holiday in the Pacific, Julie is about to send postcards to friends
in Australia, New Zealand and New Caledonia. In the local money,
stamps for Australia cost 50c, for New Zealand 60c and for New
Caledonia 80c. Half the postcards are going to Australia and the
total cost of stamps will be $14.
How many postcards is she sending?
[4 marks]
10. For full marks explain how you found your solution.
A Mexican triangle is made up of an equilateral
triangle of shaded and unshaded discs, with all
the shaded discs making up a smaller equilateral
triangle in one corner. The diagram shows a
Mexican triangle with 3 shaded discs and 12 unshaded discs. Find a Mexican triangle in which
there is more than 1 shaded disc, twice as many
unshaded discs as shaded discs, and fewer than
100 discs overall.
How many discs are there altogether, in the
Mexican triangle you have found?
[5 marks]
Western Australian
Junior Mathematics Olympiad 2011
Team Questions
45 minutes
gives 4 8 9 = 288
gives 2 8 8 = 128
gives 1 2 8 = 16
gives
1 6 = 6.
A. For the following numbers, find their fossil or show that they leave
no fossil.
(i) 273
(ii) 619
(iii) 333
(iv) 513
C.
D. Find the largest 3-digit number, all of whose digits are different,
that leaves a fossil (non zero).
E. Find the largest 3-digit number, all of whose digits are different,
that leaves an odd fossil.
F. Find the largest number, whose digits are all different, that leaves
an odd number for a fossil.
G. What proportion of 2-digit numbers leave no fossil? Try to find an
argument other than checking all the 2-digit numbers.
H. Among the 2-digit numbers, which fossil is the rarest. Explain
why.
I. Show that for every n the rarest fossil given by n-digit numbers is
the same as in H.
5. Answer: 52.
Volume of water is 100 50 40.
Volume of block is 60 40 25.
Together,
100 50 40 + 60 40 25 = 100 50 40 + 100 10 60
= 100 50 40 + 100 50 12
= 100 50 (40 + 12).
So the new depth is 40 + 12 = 52 cm.
[2 marks]
6
2
= 36
a common angle at B.
1
A
B
O
C
So we have,
AC
PC
=
CP
CB
(P C)2
AC =
= 62
CB
AB = AC + CB
= 62 + 1 = 37 m.
Note. There are at least two other ways to get the answer.
Alternative 1. In 4P CB we see P B =
since 4AP B 4P CB,
12 + 62 =
37. Then,
PB
CB
=
AB
PB
(P B)2
= 37.
AB =
CB
Alternative 2. Let O be the centre of the semi-circle, and note
that OP = OB = R, the radius of the semi-circle, and AB = 2R.
Now considering 4OCP , and noting OC = R 1 we have,
(R 1)2 + 62 = R2
R2 2R + 1 + 36 = R2
AB = 2R = 37.
[3 marks]
8. Answer: 84. Chris completes laps in 112 seconds, and Sophie is
meeting him after only 48 seconds, then he is only completing 48/112
of a lap in the time that Sophie takes to complete the rest of the
lap, namely 64/112 of a lap. Hence, Sophie is running at a speed
equal to 64/48 = 4/3 of his own speed, and will run an entire lap in
3
of his time = 43 112 = 84 seconds.
[3 marks]
4
9. Answer: 24. Say that Julie is sending c postcards to New Caledonia
and z to New Zealand. Then she is sending c + z to Australia. The
total cost will be 50(c + z) + 60z + 80c = 130c + 110z cents. This
must equal $14 = 1400 cents, so we have 130c + 110z = 1400, or
equivalently
13c + 11z = 140.
Observe that 13 11 = 2 and 10 13 = 130, so that we have
140 = 5(13 11) + 10 13
= 15 13 5 11
= (15 11t)13 + (5 + 13t)11.
Hence, the general solution over the integers is
c = 15 11t, z = 5 + 13t,
where t Z, for which only t = 1 gives c and z positive. Thus, we
find the only solution is c = 4 and z = 8 so the total number of
stamps is 4 + 8 + 12 = 24.
[4 marks]
10. Answer: 45. A triangle with n discs along the bottom contains
n(n + 1)/2 discs altogether. So if the outer triangle in a Mexican
triangle has n discs on each side and the inner triangle has m there
will be m(m + 1)/2 shaded discs and n(n + 1)/2 m(m + 1)/2
unshaded discs. We therefore have the equation,
n(n + 1)/2 m(m + 1)/2 = 2(m(m + 1)/2).
Expanding and simplifying gives
n2 + n = 3(m2 + m).
We calculate a few values of n2 + n till we find one which is 3 times
another (ignoring n = 1 since were told there is more then 1 shaded
disc, and we can stop at n = 13 since for n > 14, n(n + 1)/2 > 100):
n n2 + n
2
6
3
12
4
20
5
30
6
42
7
56
8
72
9
90
10
110
132
11
12
156
13
182
Since 90 = 3 30 we need m = 5 and n = 9, and the total number
of discs is 9 (9 + 1)/2 = 45.
For larger n, either n(n + 1) is not divisible by 3 or n(n + 1)/3 is
not in the table.
Thus the solution is 45 (and its the only solution).
[5 marks]
A. Answers: 8, 0, 4, 5.
273 7 2 7 3 = 42 7 8,
619 7 6 1 9 = 54 7 20 7 0,
333 7 3 3 3 = 27 7 14 7 4,
513 7 5 1 3 = 15 7 5.
B. Answer: 98.
99 7 9 9 = 81 7 8,
98 7 9 8 = 72 7 14 7 4.
So the largest 2-digit number that leaves a fossil of 4 is 98.
Alternatively, backtracking:
4 = 1 4 = 4 1 = 2 2.
14 = 7 2, 41 is prime, and 22 = 2 11 (11 is not a single digit).
Now 72 = 9 8 and 99 leaves a fossil of 8. So 98 is largest.
C. Answer: (i) 311 (ii) 999.
First let us discuss jargon and methodology. Thinking as in ones
family tree, one can talk of the numbers at each stage of fossilisation
as descendants. So, for
273 7 2 7 3 = 42 7 8
we have 42 and 8 as descendants of 273, with the last descendant
(8 in this case) when it is non-zero being the fossil of the number.
Thinking the other way, a word that is the opposite of descendant
is antecedent. We could say that the fossil 8 comes from 273 and
42, or that 273 and 42 are antecedents of 273.
As we discovered in B., in terms of methodology, we may work
forwards toward the fossil, i.e. find descendants, or work backwards
from the fossil, i.e. find antecedents. In working backwards, we see
that a number can have no antecedents if it is a prime of more than
one digit, or has a prime factor of more than one digit (noting that:
if a number of more than one digit is not divisible by 2, 3, 5 or 7
then it must have a prime factor of more than one digit).
F. Answer: 9751. We can use the digit 5, since we are not using any
even digits, so we are left, at best, possibly with 97531.
But 97531 7 945 7 7 0.
So we proceed by removing individual digits until we get a fossil
result. (Bracketing indicates removal of the digit.)
9753(1) behaves just like 97531.
975(3)1 7 315 7 15 7 5, a fossil!
97(5)31 7 189 7 63 7 18 7 8, a fossil!
9(7)531 7 135 7 15 7 5, again.
(9)7531 7 105 7 0.
So we have our answer:
We cant get a fossil from a 5-digit number with distinct
digits, and the only 4-digit numbers with distinct digits
that leave an odd fossil are those that use the digits 9,7,5,1
or 9,5,3,1 the largest of which is 9751.
G. Answer: 24/90 = 4/15.
The direct antecedents of 0 are 10, 20, . . . , 90 (9 of them).
Considering antecedents of 10, 20, . . . , 90:
10 [ 25, 52.
20 [ 45, 54.
30 [ 65, 56.
40 [ 85, 58.
50 to 90 have no 2-digit antecedent, since such a number would
have 5 as one digit, and at most 9 for its other digit, the product
of which is at most 45 < 50.
Thus there are 8 direct antecedents of 10, 20, . . . , 90.
Next consider the antecedents of 10, 20, . . . , 90.
25 [ 55.
52 is divisible by 13, and so has no antecedent.
45 [ 59, 95.
54 [ 69, 96.
65 is divisible by 13, and so has no antecedent.
56 [ 78, 87.
58 is divisible by 29, and so has no antecedent.
85 is divisible by 17, and so has no antecedent.
Thus we have 7 more antecedents.
Now we investigate the next level antecedents.
55 is divisible by 11, and so has no antecedent.
59 is prime, and so has no antecedent.
95 is bigger than 81, and so has no 2-digit antecedent.
69 is divisible by 23, and so has no antecedent.
96 is bigger than 81, and so has no 2-digit antecedent.
78 is divisible by 13, and so has no antecedent.
87 is bigger than 81, and so has no 2-digit antecedent.
So we have added no new 2-digit antecedents at this level and
hence we had already found all the 2-digit antecedents of 0, and in
all there are 9 + 8 + 7 = 24 out of 90.
. 1 2 3 4 5 6 7 8 9
24 1 8 2 9 6 12 2 23 3