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These questions are from the Australasian Maths Olympiad Website.

A. (Time: 3 minutes)

What is the value of:

268 + 1375 + 6179 - 168 - 1275 - 6079 ?

B. (Time: 5 minutes)

Each of 8 boxes contains one or more marbles. Each box contains a different number of marbles, except for two boxes which contain the same number of marbles. What is the smallest total number of marbles that the 8 boxes could contain?

C. (Time: 5 minutes)

Find the whole number which is:

less than 100;

a multiple of 3;

a multiple of 5;

odd, and such that,

the sum of its digits is odd.

D. (Time: 6 minutes)

Takeru has four 1 centimetre long blocks, three 5 centimetre long blocks, and three 25 centimetre long blocks. By joining these blocks to make different total lengths, how many different lengths of at least 1 centimetre can Takeru make?

E. (Time: 6 minutes)

The figure below is made up of 5 congruent squares. The perimeter of the figure is 72 cm. Find the number of square cm in the area of the figure.


A. - 300

METHOD 1: Make a simpler problem...Notice that each number being added is 100 more than one of the numbers being subtracted.

The value is 100 + 100 + 100 = 300

METHOD 2: Group by operation...Add the numbers 268 + 1375 + 6179 = 7822.

Then add the numbers 168 + 1275 + 6079 = 7522.

Finally, subtract the totals: 7822 - 7522 = 300

B. - 29 marbles

Draw a picture...

Draw 8 boxes. Then put the smallest possible number of marbles in each box. Put 1 marble in box 1. Then put 1 marble in box 2. You can't just put 1 marble in box 3, because that would make three boxes with the same number of marbles. So put 2 marbles in box 3, 3 marbles in box 4, and so on.

The smallest total number of marbles is 1+ 1 + 2 + 3 + 4 + 5 + 6 + 7 = 29 marbles.

C. - 45

Proceed one statement at a time. Eliminate those numbers which fail to satisfy all the conditions.


  • Less than 100

    1, 2, 3, ..., 99

  • Multiple of 3

    3, 6, 9, ..., 99

  • Also multiple of 5

    15, 30, 45, 75, 90

  • Odd

    15, 45, 75

  • Sum of digits is odd


D. - 79

METHOD 1: Start with a simpler problem...

(a) Lengths formed by 1 cm blocks: 1, 2, 3, 4.

(b) Lengths formed by remaining blocks: 5, 10, 15; 25, 30, 35, 40; 50, 55, 60, 65; 75, 80, 85, 90.

(c) Each of the fifteen (b) length bocks can be combined with the four (a) lengths, thus producing 15 x 4 = 60 different amounts.


(a) 4

(b) 15

(c) 60


79 different amounts


Number of choices for 1cm lengths, including 0, is 5: (0, 1, 2, 3, 4).

Number of choices for 5cm lengths, including 0, is 4: (0, 1, 2, 3).

Number of chioces for 25cm lengths, including 0, is 4: (0, 1, 2, 3).

Total number of choices for all lengths is 5 x 4 x 4 = 80. However, 80 includes the choice of having none of the lengths as a choice. Since it is given that each length must be 1cm or longer, there are 80 - 1 = 79 amounts of at least 1cm.

METHOD 3: Establish a maximum and then eliminate all impossibilities.Find that largest possible length that can be made with the blocks, and then subtract the number of smaller values that cannot be made. The maximum that can be made is 4 + 15 + 75 = 94cm. The 15 lengths less than 94cm that cannot be made are those that require four 5cm blocks. These are 20, 21, 22, 23, 24, 45, 46, 47, 48, 49, 70, 71, 72, 73 and 74cm lengths. The number of possible lengths is 94 - 15 = 79

E. - 180 cm²

Find the length of one side of the figure...

Because of the common dies, all the squares are congruent to each other. The perimeter consists of 12 equal sides. The length of a side is 72 / 12 = 6cm. The area of each square is 6 x 6 = 36cm².

The area of the figure is 5 x 36 = 180cm².

You can buy Maths Olympiad Books


by Dr George Lenchner

(Australian Edition. 2005. Reprinted with corrections 2008.)

285 pages

ISBN : 978-0-9757316-0-4


(Australian Edition. 2008.)

Editors : R. Kalman, J. Phegan, A. Prescott

320 pages

ISBN : 978-0-9757316-2-8


by Dr George Lenchner

(Australian Edition. 2006.)

290 pages

ISBN : 978-0-9757316-1-1

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โˆ™ 2011-09-13 11:34:35
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