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## Functional Equation Problem from SMO, 2018 – Question 35

Try to solve this problem number 35 from Singapore Mathematics Olympiad, SMO, 2018 based on Functional Equation.

## Problem – Functional Equation (SMO Entrance)

Consider integers ${1,2, \ldots, 10}$. A particle is initially -at 1 . It moves to an adjacent integer in the next step. What is the expected number of steps it will take to reach 10 for the first time?

• 82
• 81
• 80
• 79

### Key Concepts

Functional Equation

Equation

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Source

Challenges an Thrills – Pre – College Mathematics

## Try with Hints

First hint

If you got stuck into this problem we can start taking an expected number of steps to be $g_{n}$. We need to remember at first the particle was in 1 then it will shift to the next step so for n no of position we can expressed it as n and n -1 where n = 2,3,4,……..,100.

Now try the rest…………..

Second Hint

Now let’s continue after the last hint …………

Then $g_{n+1} = \frac {1}{2} (1+g_{n} + g_{n+1} )+ \frac {1}{2}$

which implies , $g_{n+1} = g_{n} + 2$

Now we know that,$g_{2} = 1$. Then $g_{3} = 3$, $g_{4}= 5$,………………,$g_{10}=17$

$g = g_{2}+g_{3}+g_{4}+………………..+g_{10} = 1+3+…………………+17 = 81$[ Answer]

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## Logarithm Problem From SMO, 2011 | Problem 7

Try this beautiful Logarithm Problem From Singapore Mathematics Olympiad, SMO, 2011 (Problem 7).

## Logarithm Problem From SMO

1. Let $x=\frac {1}{\log_{\frac {1}{3}} \frac {1}{2}}$+$\frac {1}{\log_{\frac {1}{5}} \frac {1}{4}}$+$\frac {1}{\log _{\frac {1}{7}} \frac{1}{8}}$ Which of the following statements
is true?
• 1.5<x<2
• 2<x<2.5
• 2.5<x<3
• 3<x<3.5
• 3.5<x<4

### Key Concepts

log function

Logarithmic

Inverse Exponentiation

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Source

Challenges and thrills – Pre – College Mathematics

## Try with Hints

First hint

If you got stuck in this problem we can start from here:

$x=\frac {1}{\log_{\frac {1}{3}} \frac {1}{2}}$+$\frac {1}{\log_{\frac {1}{5}} \frac {1}{4}}$+$\frac {1}{\log _{\frac {1}{7}} \frac{1}{8}}$

If we refer too the basic properties of log we can find ,

x=$\frac{\log \left(\frac{1}{3}\right)}{\log \left(\frac{1}{2}\right)}$+$\frac{\log \left(\frac{1}{5}\right)}{\log \left(\frac{1}{4}\right)}$+$\frac{\log \left(\frac{1}{7}\right)}{\log \left(\frac{1}{8}\right)}$=$\frac{-\log 3}{-\log 2}+\frac{-\log 5}{-\log 4}+\frac{-\log 7}{-\log 8}$

Try the rest ………………………………..

Second Hint

$\frac{-\log 3}{-\log 2}+\frac{-\log 5}{-\log 4}+\frac{-\log 7}{-\log 8}$

so we can find

$\frac {\log 3+ \log 5^{\frac {1}{2}}}+ \log 7^{\frac {1}{3}}{log 2}$

= $\frac {\log \sqrt {45} + log 7^{\frac {1}{3}}}{log 2}$ < $\frac {\log \sqrt {65} + log 8^{\frac {1}{3}}}{log 2}$

= $\frac{3 \log 2+\log 2}{\log 2}=4$

Try the rest …………………..

Final Step

Now let’s say ,

2x = $2 \frac {log 3 + log 5^{\frac {1}{2}}+ log 7^{\frac {1}{3}}}{log 2}$

=

$\begin{array}{l} \frac{\log (9 \times 5)+\log \left(49^{\frac{1}{3}}\right)}{\log 2}>\frac{\log \left(45 \times 27^{\frac{1}{3}}\right)}{\log 2} = \ \frac{\log (45 \times 3)}{\log 2}>\frac{\log (128)}{\log 2}=7 \end{array}$

so x is greater than 3.5.

3.5 <x<4 is the correct answer.

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## Problem on Probability from SMO, 2012 | Problem 33

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2012 based on Probability.

## Problem – Probability (SMO Entrance)

Two players A and B play rock – paper – scissors continuously until player A wins 2 consecutive games. Suppose each player is equally likely to use each hand – sign in every game . What is the expected number of games they will play?

• 12
• 15
• 16
• 20

### Key Concepts

Probability

Permutation Combination

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Source

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First hint

Two players are playing a series of games of Rock – Paper – scissors. There are a total of K games played. Player 1 has a sequence of moves denoted by string A and similarly player 2 has string B. If any player reaches the end of their string, they move back to the start of the string. The task is to count the number of games won by each of the player when exactly K games are being played.

To start with this particular problem let’s set an expectation as k.

If A doesnot win , so the probability will be = $\frac {no of event}{total event} = \frac {2}{3}$

so game will be restarted.

Try to find out the rest of the cases…………..

Second Hint

Continue from the 1st hint :

Again case 2: If A wins at first and then losses so again the probability will be $\frac {1}{3} \times \frac {2}{3}$. again new game will start.

Last case : A wins in consecutive two games the probability will be $\frac {1}{3} \times \frac{1}{3}$

Do the rest of the sum……..

Final Step

The total number of games that they will play :

K = $\frac {2}{3} \times (K+1) \times \frac {2}{9} (K+2) \times \frac {1}{9} \times 2$

Now solve this equation and at the end E will be 12. [Check it yourself]

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## Problem on Permutation | SMO, 2011 | Problem No. 24

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2011 based on Permutation.

## Permutation Problem (SMO Entrance)

A $4 \times 4$ Sudoku grid is filled with digits so that each column , each row and each of the four $2 \times 2$ sub grids that composes the grid contains all of the digits from 1 to 4. For example

• 288
• 155
• 160
• 201

### Key Concepts

Permutations & Combinations

Sudoku

Set Theory

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Source

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First hint

If you really get stuck in this problem here is the first hint to do that:

At 1st let’s consider the sub grids of $2 \times 2$ filled with 1-4 ( 1, 2 , 3 ,4)

If a,b,c,d are all distinct , and there are no other numbers to place in x . If {a,b} = {c,d} then again a’,b’,c,d are all distinct , and no other number can be possible for x’.

We need to understand that the choices we have ,

{a,a’} = {1,2} , {b,b’} = {3,4}, {c,c’} = {2,4} and {d,d’} = {1,3}

Among these choices $2^4 = 16$ choices 4 of them are impossible – {a,b} = {c,d} = {1,4} or {2,3} and

{a,b} = {1,4} and {c,d} = {2,3} and {a,b} = {2,3} and {c,d} = {1,4}

Try rest….

Second Hint

Now for each remaining case a’,b’,c’ and d’ are uniquely determined so

{x} = {1,2,3,4} – {a,b} $\cup$ {c,d}

{y} = {1,2,3,4} – {a,b} $\cup$ {c’,d’}

{x’} = {1,2,3,4} – {a’,b’} $\cup$ {c,d}

{y’} = {1,2,3,4} – {a’,b’} $\cup$ {c’,d’}

Final Step

In final hint :

There are 4! = 24 permutation in the left top grid we can find. So total 12 * 24 = 288 possible 4$\times$ 4 Sudoku grids can be found.

Categories

## Application of Pythagoras Theorem | SMO, 2010 | Problem 22

Try this problem from the Singapore Mathematics Olympiad, SMO, 2010 based on the application of the Pythagoras Theorem.

## Application of Pythagoras Theorem- (SMO Test)

The figure below shows a circle with diameter AB. C ad D are points on the circle on the same side of AB such that BD bisects $\angle {CBA}$. The chords AC and BD intersect at E. It is given that AE = 169 cm and EC = 119 cm. If ED = x cm, find the value of x.

• 65
• 55
• 56
• 60

### Key Concepts

Circle

Pythagoras Theorem

2D – Geometry

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Source

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First hint

If you get stuck in this problem this is the first hint we can start with :

As BE intersect $\angle {CBA}$ we have $\frac {BC}{BA} = \frac {EC}{EA} = \frac {119}{169}$

Thus we can let BC = 119 y and BA = 169 y .

Since $\angle {BCA} = 90 ^\circ$.

Then try to do the rest of the problem ………………………………………………

Second Hint

If we want to continue from the last hint we have :

Apply Pythagoras Theorem ,

$AB ^2 = AC^2 + BC ^2$

$(169y)^2 = (169 + 119)^2 + (119y)^2$

$y^2 (169-119)(169+119) = (169+119)^2$

$y^2 = \frac {169+119}{169-119} = \frac {144}{25}$

$y = \frac {12}{5}$

Final Step

In the last hint:

Hence , from triangle BCE , we have BE = $\sqrt{119^2 + (119y)^2} = 119 \times \frac {13}{5}$

Finally , note that $\triangle {ADE}$ and $\triangle {BCE}$ are similar , so we have

ED = $\frac {AE \times CE}{BE} = \frac {169 \times 119}{119 \times \frac {13}{5}} = 65$ cm .

Categories

## Functional Equations Problem | SMO, 2012 | Problem 33

Try this beautiful Problem from Singapore Mathematics Olympiad, 2012 based on Functional Equations.

## Problem – Functional equations (SMO Test)

Let L denote the minimum value of the quotient of a 3- digit number formed by three distinct divided by the sum of its digits.Determine $\lfloor 10L \rfloor$.

• 105
• 150
• 102
• 200

### Key Concepts

Functional Equation

Max and Min Value

But try the problem first…

Source

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First hint

If you got stuck at first only here is the hint to begin with :

Anyway a three digit number we can be expressed as 100x + 10 y +z depending on the place values. and if we do minimize it :

F(x y z) = $\frac {100x + 10y + z}{x + y + z}$

Lets consider that for distinct x , y , z, F(x , y , Z) has the minimum value when x<y<z.

Again we can assume,

$0 < a < b < c \leq 9$

Note ,

F(x,y,z) = $\frac {100 x + 10 y + z }{x +y + z}$ = 1 + $\frac {99 x + 9 y }{x+y+z}$

Try the rest of the sum……………

Second Hint

From the last hint we can say

F(x y z ) is minimum when c = 9 (say)

F(x y 9) = 1+ $\frac {99x +9y }{x+y+9} = 1 + \frac {9(x + y + 9) + 90 x – 81}{x+y +9} = 10 + \frac {9(10x -9)}{x+y +9}$

Try to do for the next case for minimum value when b = 8………………

Final Step

In the last hint we can do the next step which is b= 8:

F(x 8 9) = 10 + $\frac {9(10x -9)}{x+17}= 10 + \frac {90(a+17)- 1611}{x + 17} = 100 – \frac {1611}{x +17}$

which has the minimum value of x = 1 and so 10 L = 105.(answer)

Categories

## Trigonometry Problem from SMO, 2008 | Problem No.17

Try this beautiful Problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.

## Problem – Trigonometry (SMO Test)

Find the value of $(log_{\sqrt 2}(cos 20^\circ) + log_{\sqrt 2} (cos 40^\circ) + log_{\sqrt 2}(cos 80^\circ))^2$

• 32
• 15
• 36
• 20

### Key Concepts

Trigonometry

Log Function

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Source

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First hint

This one is a very simple. We can start from here :

As all are in the function of log with $\sqrt 2$ as base so we can take it as common such that

$log_{\sqrt 2}(cos 20 ^\circ . cos 40 ^\circ . cos 80^\circ)$

Now as you can see we dont know the exact value of $cos 20^\circ$ or $cos 40^\circ$ or $cos 80^\circ$ values.

But theres a formula that we can use which is

cosA.cos B = $\frac {1}{2} (cos (A+B) + cos (A-B))$

Now try apply this formula in the above expression and try to solve………

Second Hint

Now those who did not get the answer yet try this:

If we apply the formula in the expression mentioned in the last hint :

$log_{\sqrt 2}(cos 20 ^\circ . \frac {1}{2}(cos 120 ^\circ . cos 40^\circ)$

$log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{2} cos 40 ^\circ . cos 20^\circ)$

$log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{4} (cos 60 ^\circ + cos 20^\circ)$

We need to do the rest of the calculation.Try to do that …………………..

Final Step

Continue from the last hint:

$log_{\sqrt 2} \frac {1}{8} = – \frac {log_{2} 8}{log_{2}(2^{\frac {1}{2}})} = -6$

So squaring this answer = $(-6)^2 = 36$ ……………………..(Answer)

Categories

## Problem on Area of Triangle | SMO, 2010 | Problem 32

Try this beautiful problem from Singapore Mathematics Olympiad based on area of triangle.

## Problem – Area of Triangle (SMO Entrance)

Given that ABCD is a square . Points E and F lie on the side BC and CD respectively, such that BE = $\frac {1}{3} AB$ = CF. G is the intersection of BF and DE . If

$\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}$ is in its lowest term find the value of m+n.

• 12
• 19
• 21
• 23

### Key Concepts

2D – Geometry

Area of Triangle

But try the problem first…

Source

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First hint

Here is the 1st hint for them who got stuck in this problem:

At 1st we will join the points BD and CG .

Then to proceed with this sum we will assume the length of AB to be 1. The area of $\triangle {BGE}$ and $\triangle {FGC}$ are a and b respectively.

Try to find the area of $\triangle {EGC}$ and $\triangle {DGF}$……………………..

Second Hint

Now for the 2nd hint let’s start from the previous hint:

So the area of $\triangle {EGC}$ and $\triangle {DGF}$ are 2a and 2b .From the given value in the question we can say the area of $\triangle {BFC} = \frac {1}{3}$(as BE = CF = 1/3 AB\).

We can again write 3a + b = $\frac {1}{6}$

Similarly 3b + 2a = area of the $\triangle DEC = \frac {1}{3}$.

Now solve this two equation and find a and b…………..

Final Step

In the last hint :

2a + 3b = 1/3……………………(1)

3a + b = 1/6……………(2)

so in $(1) \times 3$ and in $(2) \times 2$

6a + 9b = 1

6a + 2b = 1/3

so , 7 b = 2/3 (subtracting above 2 equation)

b = $\frac {2}{21}$ and a = $\frac {1}{42}$

So,$\frac {Area of ABGD}{Area of ABCD} = 1-3(a+b) = 1- \frac {15}{42} = \frac {9}{14}$

Comparing the given values from the question , $\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}$

m = 9 and n = 14

Categories

## Problem on Functional Equation | SMO, 2010 | Problem 31

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2010 based on functional equation.

## Problem – Functional Equation (SMO Entrance)

Consider the identity $1+2+……+n = \frac {1}{2}n(n+1)$. If we set $P_{1}(x) = \frac{1}{2}x(x+1)$ , then it is the unique polynomials such that for all positive integer n,$p_{1}(n) = 1+2+…………..+n$ . In general, for each positive integer k, there is a unique polynomial $P_{k} (x)$ such that :

$P_{k} (n) = 1^k + 2^ k+3^k +………………+n^k$ for each n =1,2,3……………

Find the value of $P_{2010} (-\frac {1}{2})$ .

• 2
• 5
• 6
• 0

### Key Concepts

Polynomials

Functional Equation

But try the problem first…

Source

Challenges and Thrills – Pre College Mathematics

## Try with Hints

First hint

If you got stuck in this question we definitely can start from here:

In the question given above say k is the positive even number :

so let $f(x) = P_{k} – P_(x-1)$

Then $f(n) = n^k$ for all integer $n\geq 2$ (when f is polynomials)

Like this then $f(x) = x^k$(again for all $x \geq 2$ .

Second Hint

If you got stuck after first hint try this one

$P_{k} (-n + 1) – P_{k}(-n) = f(-n +1) = (n-1)^k$…………………………….(1)

Again, $P_{k} (-n + 2) – P_{k}(-n+1) = f(-n +2) = (n-2)^k$…………………………………(2)

Now taking n = 1;The $eq^n$(1) becomes, $P_{k}(0) – P_{k}(-1) = f(0) = 0^{k}$,

And for $eq^(n)$ (2) ; $P_{k}(1) – P_{k}(0) = f(1) = 1^{k}$.

Now sum these equation and try to solve the rest………..

Final Step

Summing this two equation we get , $P_{k}(1)-P_{k}(-n) = 1^{k} + 0^{k}+1^{k}+…….+(n-1)^k$.

so , $P_{k}(-n)+P_{k}(n-1)=0$

Again if $g(x) = (P_{k}(-x)+P_{k}(x-1)$

Then g(n) is equal to 0 for all integer $n\geq2$

As g is polynomial, g(x) =0;

So , $P_{k}(-\frac {1}{2}) + P_{k}(-\frac {1}{2}) = 0$

so $p_{k}(-\frac {1}{2}) = 0$ …………(Answer)

Categories

## Centroid of Triangle | SMO, 2009 | Problem 1

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Centroid of Triangle.

## Problem – Centroid of Triangle (SMO Entrance)

Let M and N be points on sides AB and AC of triangle ABC respectively. If $\frac {BM}{MA} + \frac {CN}{NA} = 1$ . Can we show that MN passes through the centroid of ABC?

### Key Concepts

2D – Geometry

Triangle

Menelaus’s Theorem

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Source

Challenges and Thrills – Pre College Mathematics

## Try with Hints

First hint

If we got stuck in this problem then we can start this problem by applying Menelaus’s Theorem.

It states : if a line intersects $\triangle ABC$ or extended sides at points D, E, and F, the following statement holds: $\frac {AD}{BD} \times \frac {BE}{EC} \times \frac {CF}{AF} = 1$

Again let D is the mid point of AC. As $\frac {BM}{MA} + \frac {CN}{NA} = 1$ then $\frac {CN}{NA}<1$ where N lies in the line segment CD.From the picture above we can see g is the intersection point between two lines BD and MN. So if we apply Menelaus’s Theorem we get :

$\frac {DG}{GB} . \frac {BM}{MA} .\frac {AN}{ND} = 1$

Now try the rest of the problem……………………………….

Second Hint

After the 1st hint again, $\frac {BG}{GD} = \frac{BM}{MA} . \frac {AN}{ND} = ( 1 – \frac {CN}{NA}). \frac {AN}{ND}$

= $\frac {NA – CN}{ND} = \frac {(2CD – CN) – CN}{ND}$

=$\frac {2 ND}{ND}$ = 2

Thus G is the centroid .(Proved)