Hi Brilliant!

I wrote the first Round of the South African Mathematics Olympiad today. I wanted to share the problems with the rest of you. I am eager to compare my answers.

**STAR SOLVER: Shikhar Jaiswal, Michael Tong**

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## Comments

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TopNewestA four-digit even number ABCD is formed by using four of the digits 1,2,3,4,5,6 and 7 without repetition. How many even numbers less than 2014 can be formed in this way?

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Clearly, \(A < 3\). Consider \(A = 2\). But then \(B > 0\), so it is greater than \(2014\). This leaves our one case, \(A = 1\). To make our number even, we choose \(2, 4, \) or \(6\) as the last digit. Otherwise, we can choose whatever digit we want. This gives \(3 \times 5 \times 4 = 60\).

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My solution, word-for-word

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WOW Michael! Thanks for such clear answers with explanations. I think you also deserve STAR solver.

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(a)...60

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Good!

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The answer is definitely A) 60

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Points P and Q are on line segment AB on the same side of the midpoint of AB. Point P divides AB in the ratio 2:3 and Q divides AB in the ratio 3:4. If PQ = 2, then the length of the segment AB is:

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(c)...70 looks correct here

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Yeah, I also got that.

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\(|\frac{4}{7} - \frac{3}{5}|x = 2 \rightarrow x = 70\)

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Rectangle ABCD has sides AB = 5 and CB = 3. Points E and F lie on diagonal AC such that AE = EF = FC. What is the area of triangle BEF?

A 3/2

B 5/8

C 5/2

D \((1/2) *\sqrt{34}\)

E \((1/3) *\sqrt{68}\)

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(c)...5/2...cause the 3 triangles have equal areas(length of altitude and base are same)

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Correct! I've awarded you the STAR SOLVER (see note description).

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C)5/2 because the there are 3 same triangles which means same area.

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Note that \([BAE] = [BEF] = [BFC]\), since they have equal bases and the same altitude. Thus, \([BEF] = \frac{1}{6} [ABCD] = \frac{5}{2}\)

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A right angled triangle has an area of 5. The altitude perpendicular to the hypotenuse has a length of 2. The perimeter of the triangle is:

A \(5\sqrt{5} + 5\)

B CUBEROOT(3) + 3

C \(5\sqrt{3} + 3\)

D \(3\sqrt{5} + \)\(5\sqrt{3} \)

E \(3\sqrt{5} + 5\)

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(E)....Take out length of hypotenuse using the information it comes out to be 5 units

Let one side be \(a\)

the other side=\(\sqrt{25-a^2}\)

Given:

\(\frac {1}{2}a\sqrt{25-a^2}=5\)

On solving...\(a=2\sqrt{5}\)

while the other side=\(\sqrt{5}\)

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I got the same answer!

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When the tens digit of a three digit number "abc" is deleted, a two-digit number "ac" is formed. How many numbers "abc" are there such that "abc" = 9 x "ac" + 4c. For example, 245 = 9 x 25 + 4 x 5

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(d)..6

The numbers are-155,515,245,425,335,605

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I got 6 (i.e. option D) as my answer. The numbers I found were: 155, 245, 335, 425, 515 and 605.

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The details of the problem are useless. We just need the algebra. \(100a + 10b + c = 9(10a + c) + 4c\)

\(100a + 10b + c = 90a + 13c\)

\(10(a+b) = 12c\)

Thus \(c = 5\). \(a+b = 6\), since \(a \neq 0\) we get \(6\) solutions, namely \(155, 245, 335, 425, 515, 605\).

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It is no details. It is just an example!

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Three standard dice are rolled and the numbers thrown are added. The probability of getting a sum of 15 is:

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Note the bijection between rolls that sum to \(15\) and rolls that sum to \(6\).

Thus, we count the probability of getting a \(6\): Since each die must be at least one, we can count instead the number of ways to put \(3\) balls in \(3\) urns, namely \({5 \choose 3} = 10\).

The total number of outcomes (assuming distinct dice) is \(6^3 = 216\).

The desired probability is \(\frac{5}{108}\)

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(d)...5/108 looks like the correct one..

Total possibilities=6^3=216

Favourable possibilities:-(3,6,6)-'3" from here

no of favourable possibilities=10

Probability=10/216=5/108

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Yeah, that's correct!

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