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**

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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?

Log in to reply

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 \(3 \times 5 \times 4 = 60\).

Log in to reply

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

Log in to reply

My solution, word-for-word

Log in to reply

The answer is definitely A) 60

Log in to reply

(a)...60

Log in to reply

Good!

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

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:

Log in to reply

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

Log in to reply

(c)...70 looks correct here

Log in to reply

Yeah, I also got that.

Log in to reply

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

Log in to reply

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 \(\frac{5}{108}\)

Log in to reply

(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

Log in to reply

Yeah, that's correct!

Log in to reply

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

Log in to reply

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 \(155, 245, 335, 425, 515, 605\).

Log in to reply

It is no details. It is just an example!

Log in to reply

(d)..6

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

(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=\(\sqrt{5}\)

Log in to reply

I got the same answer!

Log in to reply