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## 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

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

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………

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 …………………..

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)

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## Problem on HCF | SMO, 2013 | Problem 35

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2013 based on HCF.

## Problem on HCF | SMO, 2013 | Problem 35

What is the smallest positive integer n,where $n \neq 11$ such that the highest common factor of n-11 and 3n +20 is greater than 1?

• 62
• 65
• 66
• 60

### Key Concepts

HCF and GCD

Number Theory

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

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

Let d is the highest common factor of n-11 and 3n +20 which is greater than 1.

So d|(n-11) and d|(3n + 20) .

If we compile this two then d|(3n +20 -3(n-11) when d|53 .

Now one thing is clear that 53 is a prime number and also d >1

so we can consider d = 53.

Now try the rest………………

Now from the previous hint

n-11 = 53 k (let kis the +ve integer)

n = 53 k +11

So for any k 3n +20 is a multiple of 53.

so 3n + 20 = 3(53k +11) +20 = 53(3k+1)

Finish the rest………..

Here is the final solution :

After the last hint :

n = 64 (if k = 1) which is the smallest integer. as hcf of (n – 11,3n +20)>1(answer)

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## 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.

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• 19
• 21
• 23

### Key Concepts

2D – Geometry

Area of Triangle

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

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 $\triangle {DGF}$……………………..

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…………..

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

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## 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 $P_{2010} (-\frac {1}{2})$ .

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• 5
• 6
• 0

### Key Concepts

Polynomials

Functional Equation

Challenges and Thrills – Pre College Mathematics

## Try with Hints

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 $x \geq 2$ .

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………..

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)

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## 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

Challenges and Thrills – Pre College Mathematics

## Try with Hints

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……………………………….

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)

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## Series Problem | SMO, 2013 | Problem 27

Try this beautiful problem from Singapore Mathematics Problem, SMO, 2012 based on Series. You may use sequential hints to solve the problem.

## Series Problem – (SMO Entrance)

Find the value of $\lfloor(\frac {3+\sqrt {17}}{2})^{6} \rfloor$

• 2041
• 1504
• 1699
• 2004

### Key Concepts

Series

Algebra

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

If you got stuck in this hint please do through this hint :

We can assume two integers, $\alpha$ and $\beta$ where ,

$\alpha = \frac {3+\sqrt {17}}{2}$ so let $\beta = \frac {3 – \sqrt {17}}{2}$

Then if we try to substitute the value in any formula lets try to find the value of $\alpha . \beta$

and $\alpha + \beta$

I think you can definitely find the values………Try …………………..

If you are still not getting it ………………

$\alpha .\beta = \frac {3+\sqrt {17}}{2} . \frac {3-\sqrt {17}}{2} = -2$

and again $\alpha + \beta = 3$

Lets consider the sum $s_{n} = \alpha ^ {n} + \beta ^{n}$.

Then try to find the value of $3 s_{n+1} + 2 s_{n}$ …………………………..

This one is the last hint ………..

$3 s_{n+1} + 2 s_{n} = (\ (alpha +\beta)(\alpha^{n+1} + \beta^ {n+1} -\alpha\beta(\alpha^{n} + \beta ^{n}$

= $\alpha ^ {n+2} – \beta ^ {n+2} = s_{n+2}$

Note that $|\beta| < 1$ .Then for even positive integer n, $\lfloor a^{n}\rfloor = s_{n} + \lfloor – \beta^n \rfloor = s_{n} – 1$.

As $s_{0} = 2$ and $s_{1}=3$ we can find $s_{6} = 2041$

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## Problem on Prime Numbers | SMO, 2012 | Problem 20

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

## Problem on Prime Numbers – (SMO Test)

Let A be a 4 – digit integer. When both the first digit (leftmost) and the third digit are increased by n, and the second digit and the fourth digit are decreased by n, the new number is n times A. Find the value of A.

• 1201
• 1551
• 1818
• 2000

### Key Concepts

Algebra

Prime Number

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

If you got stuck you can follow this hint:

We can assume the 4 digit number to be A = $\overline {abcd}$

If we expand it into the equation

1000(a+n) + 100(b – n) + 10(c+n) + (d-n) = nA

Try the rest of the sum ………..

After the previous hint :

If we compare the equation it gives :

A + 909 n = nA or

(n-1)A = 909 n

Now one thing we can understand that n and (n-1) are relatively prime and 101 is a prime number . So n= 2 or n= 4.

We have almost got the answer .So try to do the rest now ……….

If n = 4 then A = 1212, which is impossible right?

as b<n given .so

n=2 and A = $909 \times 2$ = 1818

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## Problem on Series | SMO, 2009 | Problem No. 25

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Problem on Series.

## Problem on Series | SMO Test

Given that $x+(1+x)^2+(1+x)^3+………(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$ where each $a_r$ is an integer , $r = 0,1,2,…,n$

Find the value of n such that $a_0 +a_1 +a_2+a_3 +………..+a_{n-2}+a_{n-1} = 60 -\frac{n(n-1)}{2}$?

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• 5
• 6
• 0

### Key Concepts

Series Problem

Algebra

Challenges and Thrills – Pre – college Mathematics

## Try with Hints

If you got stuck in this sum then we can try by understanding the pattern we are using here.If we assume x = 1 thenn the expression will be

$x+(1+x)^2+(1+x)^3+………(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$

The right hand side of this equation will be

$a_0 +a_1 + a_2 +a_3+………………+ a_n$

and the left hand side of the given equation be like

$1 + 2^2 + 2^ 3 + ……+2^n$

$1 + 2^2 + 2^ 3 + ……+2^n$ = $a_0 +a_1 + a_2 +a_3+………………+ a_n$

Try the rest of the sum…………..

In the next hint we continue from the previous hint:

so the expression , $1 + 2^2 + 2^ 3 + ……+2^n$ = $2^{n+1} – 3$

Again , $a_1 = 1+2+3+……….+n = \frac {n(n+1)}{2}$

so $a_n = 1$

Now we have almost reach the final step.I am not showing it now.Try first………

Now coming back to the last step :

$60 – \frac {n(n+1)}{2} + \frac {n(n+1}{2} + 1 = (2^{n+1} – 3$

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## Digit Problem from SMO, 2012 | Problem 14

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

## Digit Problem – Cubic Equation – SMO Test

Find the four digit number $\overline{abcd}$ satisfying

$2 \overline {abcd} + 1000 = \overline {dcba}$

• 2996
• 2775
• 1611
• 2001

### Key Concepts

Algebra

Digit Problem

Cubic Equation

Challenges and Thrills – Pre – college Mathematics

## Try with Hints

If you got stuck with this sum lets try to start from here :

Let us consider the left hand side of the given equation –

$2 \overline {abcd} + 1000$ . Here if we express it in equation

suppose $n_{1} = ax^3 +bx^2+cx+d = \overline {abcd}$

$n_{0} = 1x^3 +0x^2+0x+0 = 1000$

$n_{0} = dx^3 +cx^2+bx+a = \overline {dcba}$ = which is the right hand side of the given equation.

So the equation becomes = > $2 n_{1} + 1000 = n_{2}$

Now compare the coefficients and try to do the problem ……………….

Lets continue after hint 1 we get :

2a + 1 = d

2b = c

2c = b

2d = a

from this we can understand that $2a +1 \leq d \leq 9$ as d is th digit number.

Now as 2d = a so a should be even number. Then the limits will be {2,4}.

Now implement the values of a and d and try to find the answer.

Here is the rest of the problem:

If we consider a = 4

Then $d \geq 2a+1 = 9$ ; thus d = 9. But $2d = a \neq 8$.Its a contradiction.

So again , a = 2

$d \leq 2a+1 =5$ ; thus 2d = a = 2

d = 6; So d + d = 6+6 which can be written as 2 carry 1.

So, a = 2 and b = 6. Now the carry over remaining equation is

b c

b c

0 1

is equal to 1 c b

so either 2c +1 = b and 2b = 10 + c

r,

2c +1 = 10 + b and 2 b + 1 = 10 + c so b = c = 9

$\overline {dcba} = 2996$(Answer)

Categories

## What we say is definition:

Sequence : A sequence is an arrangement of objects or a set of numbers in a particular order followed by some rule . In other words we can say that each sequence has a definite pattern. For example :

Example 1 : {1,2,3,4,5,…………………………} – here if we add 1 with the previous term then we are getting the next term as 1 , 1+1 = 2 , 2+1 = 3, and so on.

Again in a sequence the terms can repeat itself such as :

{0,1, 0, 1 , 0 , 1 ,……………} – here 1’s and 0’s are alternately repeating itselves.

Series : A “series” is what you get when you add up all the terms of a sequence; the addition, and also the resulting value, are called the “sum” or the “summation”.

For an example if we say there is a sequence of {1,2,3,4} then the corresponding series is {1+2+3+4} and the sum of this series is 10.

### Know Something More About Sequence :

In a sequence each number is called TERM or ELEMENT or MEMBER .

Sequences can be of two types (primarily ) :

(1) Finite Sequences : These are the sequences where the last term is defined in other words. We can say it has a finite number of terms . For an example we can say :

{1,2,3,4,5} – here the last term is already defined so this is a finite sequence .

{4,3,2,1} – We can apply the same logic and can say this is a finite sequence as well (only its in backward )

(2) Infinite Sequences : Thee are the sequences where the last term is not defined .In other words we can say it has an infinite number of terms. For example :

{1,2,3,4,…………………….} – here we have used some dots after 4 instead of any number . The only reason for this is to tell you it can continue till infinity. Huh! funny…….

For this reason these types of sequences are called infinite sequences.

Apart from these two there are some commonly used sequences we have :

1. Arithmetic Sequences: In these sequences every term is created by adding or subtracting a definite number to the preceding number. Example : {1,5,9,13,17,21,25,…} – where the difference of (5-1) = 4 , (9-5) = 4 and so on…
2. Geometric Sequences : In these sequences every term is obtained by multiplying or dividing a definite number with the preceding number. Example : { 6, 12, 24, 48 ,…} -where if we divide the next term by the previous term then $\frac {12}{6} = 2$ again $\frac {24}{12} = 2$ and so on………………….

## Some examples for better understanding :

Before starting with an example lets try to find the importance of formula to represent one sequence :

Now from this sequence we can understand that

1st term is = 3

2nd term is = 5

3rd term is = 7

4th term is = 9 and so on .

So if I tell you to find the 10 th term (lets say each term has a general name which is ‘n’) of this sequence then it will be easy for us to find i.e we can continue counting the terms and we can say the 10th term is 21 (HUH – that’s easy) but if I tell you to find the 100th term from this sequence then ???????????????????

Its not impossible to find but it will be a waste of time , page , ink and energy. For this if we can generate a formula from one sequence we can work at ease.

So from the above sequence {3,5,7,9,…………….}

Lets draw a table and lets start considering n as the general formula for the given sequence:

[Note = We have to match the (we want to get) column and the (reality) column ]

Now again considering the formula as 2n such that :

So gain the two columns are not matching but one thing we say that the gaps between two terms are same as given in the sequence {3,5,7,9,………………………………}.So we are not far from the correct answer.

Now its perfectly matches with the columns. So the desire formula of the sequence is {3,5,7,9, …..} = 2n + 1.

I hope we can generate some more formula with this method. Try to do ……..

## Sequence Problems :

Calculate 4th term of the sequence :

$a_{n} = (-n)^{n}$

$a_{1} = – 1^{1}$ = -1

$a_{2} = (- 2 )^{2}$ = 4

$a_{3} = (- 3 )^{3}$ = -27

$a_{4} = (- 4 )^{4}$ = 256 (Answer )

For the sequence defined by $a_{n} = n^{2} – 5n + 2$ , what is the smallest value of n for which  $a_{n}$ is positive ?

$a_{n} = n^{2} – 5n + 2$

Therefore ,

$a_{1} = 1^{2} – 5$ times $1 + 2 = 1 – 5 +2 = -2 < 0$

$a_{2 } = 2^{2} – 5$ times $2 + 2 = 4 – 10 +2 = -4 < 0$

$a_{3} = 3^{2} – 5\times 3 + 2 = 9 – 15 +2 = -4 < 0$

$a_{4} = 4^{2} – 5\times 4 + 2 = 16 – 20 +2 = -2 < 0$

$a_{5} = 5^{2} – 5\times 5 + 2 = 25 – 25 +2 = 2 > 0$

Therefore the smallest value of n for which $a_{n}$ is positive is n = 5 .