Question 1

What is the sum of the first five square numbers ending in a 1?

Solution 1

Note that for a square number to end in a 1, its square root must end in a 1 or a 9. So the sum of the first five square numbers ending in a 1 is:

12 + 92 + 112 + 192 + 212 = 1 + 81 + 121 + 361 + 441 = 1005

Question 2

In the diagram, PR = QR, ÐPRQ = 40^°, and ÐPTU = 25^°. What is ÐRST, in degrees?

Solution 2

Since PR = QR, the triangle PQR is isosceles and ÐRPQ = ÐRQP. Considering the angles in triangle PQR gives the equation ÐRPQ + ÐRQP = 180^° -ÐPRQ = 180^° - 40^° = 140^°. Therefore, ÐRPQ = ÐRQP = 70^° and ÐSQT = ÐRQT = 180^° - ÐPQR = 180^° - 70^° = 110^°. Considering the angles in triangle SQT gives us ÐQST = 180^° - ÐSQT - ÐSTQ = 180^° - 110^° -25^° = 45^°. Hence, ÐRST = 180^° - ÐQST = 180^° - 45^° = 135^°.

Question 3

Simplify

16Ö7 Ö7 +  1 Ö7

Solution 3

16Ö7 Ö7+  1 Ö7 =  16Ö7 Ö7+  1 Ö7 ×  Ö7 Ö7 =  16 ×7 7 + 1 =  102 8 = 14

Question 4

Mary has sat for 10 tests this term and has an average score of 68. What mark must she gain in the next one to raise her average to 70?

Solution 4 Let x denote the mark that Mary must receive in the next test. In the previous 10 tests, the total number of marks that Mary received was 68 ×10 = 680. To raise her average to 70 in the next test, the following equation must hold:

680 + x 11 = 70 Þ 680 + x = 70 ×11 = 770 Þ x = 770 - 680 = 90

Question 5

The digits 3, 1, 4 and 1 can be arranged to form how many different 4-digit numbers?

Solution 5 If we treat the two 1's as two different numbers, then there would be 4! = 24 ways of rearranging the four digits. (There would be one of four different numbers which we could choose for the thousands digit, one of three numbers for the hundreds digit, one of two numbers for the tens digit, and only one choice for the units digit.) However, since the two 1's can be interchanged, we have counted each number twice. Therefore, there are only [ 24/2] = 12 different 4-digit numbers that can be made from the digits 3, 1, 4 and 1.

Question 6

What is the surface area, in square centimetres, of a cube having volume 343 cm³?

Solution 6 Since the cube has a volume of 343 cm³, the side length of the cube must be ³Ö{343} cm = 7 cm. Thus, the area of one square face of the cube is 7 cm ×7 cm = 49 cm². The surface area of the cube is composed of six of these square faces and is 6 ×49 cm² = 294 cm².

Question 7

What is the area of the smallest circle in which you can fit six equilateral triangles of area 3 without overlap? (Express your answer exactly)

Solution 7 If we inscribe a regular hexagon in a circle and draw its diagonals through the centre, we can see that in a circle of radius r, we can exactly fit six equilateral triangles of side length r. For an equilateral triangle to have area 3, we require the following equation to hold:

Area =  1 2 r2 sin60° = 3  1 2 r2 ×  Ö3 2 = 3 r2 =  12 Ö3 r2 = 4 Ö3

Thus the area of the required circle is pr² = 4 Ö3p.

Question 8

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.

Solution 8 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:

80t = 100(t - 0.5) 80t = 100t - 50 20t = 50 t =  50 20 = 2.5

Therefore, Luke is arrested 2.5 hours after 8:00am - that is, at 10:30 am.

Question 9

Norm and Geordie each roll a die. What is the probability that the product of the two numbers rolled is less than 6? (Express your answer as a fraction in simplest form.)

Solution 9 Let the number that Norm rolls be N and the number that Geordie rolls G. Since there are six possibilities for N and for G, all equally likely, there are 6 ×6 = 36 possibilities for the pair (N, G). The only possible pairs whose product is less than 6 are (1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (1, 5), (5, 1) and (2, 2).

Since there are 10 of these pairs, the required probability is [ 10/36] = [ 5/18].

Question 10

Three jolly professors, Tim, Swarup and Kris, are gambling by a billabong. They start with sums of money in the ratio 7:6:5 and finish with sums of money in the ratio 6:5:4, in the same order of wealth. One of the professors won $12. How many dollars did he start with?

Solution 10 Suppose that the total amount of money of all the professors in dollars is M. Since they started with sums of money in the ratio 7:6:5, they each started with [ 7M/(7+6+5)], [ 6M/(7+6+5)] and [ 5M/(7+6+5)], respectively. Since they ended with sums of money in the ratio 6:5:4, they each ended with [ 6M/(6+5+4)], [ 5M/(6+5+4)] and [ 4M/(6+5+4)], respectively. Thus, the first professor won [ 6M/15] - [ 7M/18] = [ M/90], the second professor won [ 5M/15] -[ 6M/18] = 0 and the third professor lost [ 5M/18] -[ 4M/15] = [ M/90]. We are told that one of the professors won $12, so M = 90 ×$12 = $1080. Hence, he started with [ 7M/18] = [ 7/18] ×$1080 = $420.

Question 11

Many years ago Melbourne University had under 5000 students enrolled. A third of the students were in first year, two-sevenths were in second year, one-fifth were in third year and the rest were postgraduate students. The mathematics department offered a popular course in which were registered a fortieth of all the first-year students, a sixteenth of all the second-year students, and a ninth of all the third-year students, while the remaining third of the maths class were all postgraduates. How many students were there in the maths class?

Solution 11

Let N denote the total number of students enrolled and let C denote the number of students registered in the maths class. Then [ 1/3] ×[ 1/40] ×N = [ N/120] first year students are in the maths class, [ 2/7] ×[ 1/16] ×N = [ N/56] second year students are in the maths class, and [ 1/5] ×[ 1/9] ×N = [ N/45]. N must be divisible by 120, 56 and 45 and hence, N is divisible by the lowest common multiple of 120, 56 and 45 - that is, 2520. But the only multiple of 2520 which is less than 5000 is 2520 itself so N = 2520. Since first, second and third year students make up two-thirds of the maths class, we have the equation:

N 120 +  N 56 +  N 45 =  2C 3 Þ  2C 3 =  2520 120 +  2520 56 +  2520 45 = 21 + 45 + 56 = 122 Þ C =  3 2 ×122 = 183

Question 12

In the diagram, the lengths of the sides of the triangle are 8, 9, 13 centimetres . The centres of the circles are at the vertices of the triangle, and the circles just touch. What are the radii of the three circles, in increasing order?

Question 13

Lines from the vertices of a square to the midoints of the sides are drawn, as shown. If the area of the large square is 1, what is the area of the smaller square in the middle?

Question 15

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( ¼f(2¹⁶)))) where f is applied 10 times?

Solution 15

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(¼f(2¹⁶)¼))) = 4 if the function f is applied five or more times.

Question 16

How many points with positive integer coordinates are there strictly inside the area bounded by the lines x=0, y=0 and the graph y = 10/x?

Solution 16 A point (x,y) will lie strictly inside the area bounded by the lines x = 0, y = 0 and the graph y = 10/x if xy < 10, where x and y are positive. The only pairs (x,y) of integers which satisfy this are the pairs:

• (1,k) with 1 £ k £ 9
• (k,1) with 1 £ k £ 9
• (2,2), (2,3), (3,2), (2,4), (4,2) and (3,3)

Question 17

Solve the following equation for the integer n:

3 æ Ö     n + Ö n2 + 8   + 3 æ Ö     n - Ö n2 + 8   = 8

Solution 17 Let a = ³Ö{n + Ö{n² + 8}} and b = ³Ö{n - Ö{n² + 8}}. Then the equation given to us is a + b = 8. We can also find the product of a and b.

ab = 3 æ Ö     n + Ö n2 + 8   3 æ Ö     n - Ö n2 + 8   = 3 æ Ö     (n + Ö   n2 + 8   ) ( 3 æ Ö     n - Ö n2 + 8   )   = 3 Ö   n2 - (n2 + 8)   = 3 Ö   -8   = -2

Now we note that:

2n = (a3 + b3) = (a + b)3 - 3ab(a+b) = 83 - 3 ×(-2) ×8 = 512 - (-48) = 560

Therefore, n = 280.

Question 18

On the planet Dankuhn lives a being which can be one of three sex types: male, female and emale. Any two different sexes may breed and the offspring from such a union is of the third sex. How many of an emale's great great great great grandparents were emales?

Solution 18 The following table shows the number of emales, females and males for each generation in the past six generations of an emale. So we can see that every emale has 22 great great great great grandparents.

	1	2	3	4	5	6
emales	0	2	2	6	10	22
females	1	1	3	5	11	21
males	1	1	3	5	11	21

Question 19

Find the number of three digit numbers whose digit sum is ten.

Solution 19 First, we will determine all integer triples (a, b,c) with 0 £ a £ b £ c £ 9 where a + b + c = 10. Note that a = 0, 1, 2 or 3.

• If a = 0, then b + c = 10, so the only possible triples are (0,1,9), (0,2,8), (0,3,7), (0,4,6) and (0,5,5).
• If a = 1, then b + c = 9, so the only possible triples are (1,1,8), (1,2,7), (1,3,6) and (1,4,5).
• If a = 2, then b + c = 8, so the only possible triples are (2,2,6), (2,3,5) and (2,4,4).
• If a = 3, then b + c = 7, so the only possible triple is (3, 3, 4).

Question 20

``Baker's Dozen" doughnuts are sold only in boxes of 7, 13 or 25. To buy 14 doughnuts you must order two boxes of 7, but you cannot buy exactly 15 since no combination of boxes contains 15 doughnuts. What is the largest number of doughnuts that cannot be ordered using combinations of these boxes?

Solution 20 Note that we can buy any number of doughnuts from 45 through to 51 as follows:

45 = 1 ×7 + 1 ×13 + 1 ×25 46 = 3 ×7 + 0 ×13 + 1 ×25 47 = 3 ×7 + 2 ×13 + 0 ×25 48 = 5 ×7 + 1 ×13 + 0 ×25 49 = 7 ×7 + 0 ×13 + 0 ×25 50 = 0 ×7 + 0 ×13 + 2 ×25 51 = 0 ×7 + 2 ×13 + 1 ×25

Since these are seven consecutive numbers, we can buy any number of doughnuts which is 45 or greater, by buying extra boxes of 7. However, it is not possible to buy 44 doughnuts.