Categories

## Triangle Problem | PRMO-2018 | Problem No-24

Try this beautiful Trigonometry Problem based on Triangle from PRMO -2018, Problem 24.

## Triangle Problem – PRMO 2018- Problem 24

If $\mathrm{N}$ is the number of triangles of different shapes (i.e. not similar) whose angles are all integers (in degrees), what is $\mathrm{N} / 100$ ?

,

• $15$
• $22$
• $27$
• $32$
• $37$

Trigonometry

Triangle

Integer

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

Prmo-2018, Problem-24

#### Check the answer here, but try the problem first

$27$

## Try with Hints

#### First Hint

Given that $\mathrm{N}$ is the number of triangles of different shapes. Therefore the different shapes of triangle the angles will be change . at first we have to find out the posssible orders of the angles that the shape of the triangle will be different…

Now can you finish the problem?

#### Second Hint

case 1 : when $x \geq 1$ & $y \geq 3 \geq 1$
$$x+y+z=180$$
$={ }^{179} \mathrm{C}_{2}=15931$
Case 2 : When two angles are same
$$2 x+y=180$$
1,1,178
2,2,176
$\vdots$
89,89,2

#### Solution

But we have one case $60^{\circ}, 60^{\circ}, 60^{\circ}$
$$\text { Total }=89-1=88$$
Such type of triangle $=3(88)$
When 3 angles are same $=1(60,60,60)$
So all distinct angles’s triangles
$$\begin{array}{l} =15931-(3 \times 88)-1 \ \neq 3 ! \ =2611 \end{array}$$
Now, distinct triangle $=2611+88+1$
$=2700 \ N=2700 \ \frac{N}{100}=27 \$

Categories

## Value of Sum | PRMO – 2018 | Question 16

Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.

## Value of Sum – PRMO 2018, Question 16

What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$

• $50$
• $53$
• $55$
• $59$
• $65$

### Key Concepts

Odd-Even

Sum

integer

Answer:$55$

PRMO-2018, Problem 16

Pre College Mathematics

## Try with Hints

We have to find out the sum . Now substitite $i=1,2,3…9$ and observe the all odd-even cases……

Can you now finish the problem ……….

$i=1 \Rightarrow$$1+(2+4+6+8+10-3-5-7-9) =1-4+10=7 i=2 \Rightarrow$$0 \times 2+(3+5+7+9-4-6-8-10)$
$=-4$

$i=3 \Rightarrow$$1 \times 3+(4+6+8+10-5-7-9) =3-3+10=10 i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$
$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10 =13 i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$
$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16 i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$
$i=9 \Rightarrow$$1 \times 9+(10)=19 Can you finish the problem…….. Therefore S =(7+10+13+16+19)-(4-3-2-1) =55 ## Subscribe to Cheenta at Youtube Categories ## Chessboard Problem | PRMO-2018 | Problem No-26 Try this beautiful Chessboard Problem based on Chessboard from PRMO – 2018. ## Chessboard Problem – PRMO 2018- Problem 26 What is the number of ways in which one can choose 60 units square from a 11 \times 11 chessboard such that no two chosen square have a side in common? , • $56$ • $58$ • $60$ • $62$ • $64$ ### Key Concepts Game problem Chess board combination ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-26 #### Check the answer here, but try the problem first $62$ ## Try with Hints #### First Hint Total no. of squares =121 Out of these, 61 squares can be placed diagonally. From these any 60 can be selected in { }^{61} C_{60} ways =61 Now can you finish the problem? #### Second Hint From the remaining 60 squares 60 can be chosen in any one way Total equal to { }^{61} \mathrm{C}{60}+{ }^{60} \mathrm{C}{60}=61+1=62 ## Subscribe to Cheenta at Youtube Categories ## Measure of Angle | PRMO-2018 | Problem No-29 Try this beautiful Trigonometry Problem based on Measure of Angle from PRMO -2018. ## Measure of Angle – PRMO 2018- Problem 29 Let D be an interior point of the side B C of a triangle ABC. Let l_{1} and l_{2} be the incentres of triangles A B D and A C D respectively. Let A l_{1} and A l_{2} meet B C in E and F respectively. If \angle B l_{1} E=60^{\circ}, what is the measure of \angle C l_{2} F in degrees? , • $25$ • $20$ • $35$ • $30$ • $45$ ### Key Concepts Trigonometry Triangle Angle ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-29 #### Check the answer here, but try the problem first $30$ ## Try with Hints #### First Hint According to the questations at first we draw the picture . We have to find out the value of \angle C l_{2} F. Now at first find out $\angle AED$ and $\angle AFD$ which are the exterioe angles of $\triangle BEL_1$ and $\triangle CL_2F$. Now sum of the angles is 180^{\circ} Now can you finish the problem? #### Second Hint \angle E A D+\angle F A D=\angle E A F=\frac{A}{2} \angle A E D=60^{\circ}+\frac{B}{2} \angle A F D=\theta+\frac{C}{2} Therefore \quad \ln \Delta A E F: \frac{A}{2}+60^{\circ}+\frac{B}{2}+\theta+\frac{C}{2}=180^{\circ} 90^{\circ}+60^{\circ}+\theta=180^{\circ} (as sum of the angles of a Triangle is 180^{\circ} Therefore \quad \theta=30^{\circ} ## Subscribe to Cheenta at Youtube Categories ## Good numbers Problem | PRMO-2018 | Question 22 Try this beautiful good numbers problem from Number theory from PRMO 2018, Question 22. ## Good numbers Problem – PRMO 2018, Question 22 A positive integer k is said to be good if there exists a partition of {1,2,3, \ldots, 20} into disjoint proper subsets such that the sum of the numbers in each subset of the partition is k. How many good numbers are there? • 4 • 6 • 8 • 10 • 2 ### Key Concepts Number theorm good numbers subset ## Check the Answer Answer:6 PRMO-2018, Problem 22 Pre College Mathematics ## Try with Hints What is good numbers ? A good number is a number in which every digit is larger than the sum of digits of its right (all less significant bits than it). For example, 732 is a good number, 7>3+2 and 3>2 . Given that k is said to be good if there exists a partition of {1,2,3, \ldots, 20} into disjoint proper subsets such that the sum of the numbers in each subset of the partition is k. Now at first we have to find out sum of these integers {1,2,3, \ldots, 20}. Later create some partitions such that two partitions be disjoint set and sum of the numbers of these partitions be good numbers Can you now finish the problem ………. Sum of numbers equals to \frac{20 \times 21}{2}=210 \& 210=2 \times 3 \times 5 \times 7 So \mathrm{K} can be 21,30,35,47,70,105 Can you finish the problem…….. Case 1 : \mathrm{A}=\{1,2,3,4,5,16,17,18,19,20\}, \mathrm{B}=\{6,7,8,9,10,11,12,13,14,15\} Case 2 : A=\{20,19,18,13\}, B=\{17,16,15,12,10\}, C=\{1,2,3,4,5,6,7,8,9,11,14\} Case 3 : \mathrm{A}=\{20,10,12\}, \mathrm{B}=\{18,11,13\}, \mathrm{C}=\{16,15,9,2\}, \mathrm{D}=\{19,8,7,5,3\}, \mathrm{E}=\{1,4,6,14,17\} Case 4 : A=\{20,10\}, B=\{19,11\}, C=\{18,12\}, D=\{17,13\}, E=\{16,14\}, F=\{1,15,5\}, G=\{2,3,4,6,7,8\} Case 5 : A=\{20,15\}, B=\{19,16\}, C=\{18,17\}, D=\{14,13,8\}, E=\{12,11,10,2\}, F=\{1,3,4,5,6,7,9\} Case 6 : A=\{1,20\}, B=\{2,19\}, C=\{3,18\} \ldots \ldots \ldots \ldots, J=\{10,11\} Therefore Good numbers equal to 6 ## Subscribe to Cheenta at Youtube Categories ## Polynomial Problem | PRMO-2018 | Question 30 Try this beautiful Polynomial Problem from Number theorm from PRMO 2018, Question 30. ## Polynomial Problem – PRMO 2018, Question 30 Let P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{n} x^{n} be a polynomial in which a_{i} is non-negative integer for each \mathrm{i} \in{0,1,2,3, \ldots, \mathrm{n}} . If \mathrm{P}(1)=4 and \mathrm{P}(5)=136, what is the value of \mathrm{P}(3) ? • 30 • 34 • 36 • 39 • 42 ### Key Concepts Number theorm Polynomial integer ## Check the Answer Answer:34 PRMO-2018, Problem 30 Pre College Mathematics ## Try with Hints Given that P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{n} x^{n} where \mathrm{P}(1)=4 and \mathrm{P}(5)=136. Now we have to find out P(3). Therefore if we put x=1 and x=5 then we will get two relations . Using these relations we can find out a_0 , a_1, a_2 . Can you now finish the problem ………. a_{0}+a_{1}+a_{2}+\ldots \ldots+a_{n}=4 \Rightarrow a_{i} \leq 4 a_{0}+5 a_{1}+5^{2} a_{2}+\ldots+a 5^{n} a_{n}=136 \Rightarrow a_{0}=1+5 \lambda \Rightarrow a_{0}=1 Can you finish the problem…….. Hence 5 a_{1}+5^{2} a_{2}+\ldots \ldots+5^{n} a_{n}=135 a_{1}+5 a_{2}+\ldots 5^{n-1} a_{n-1}=27 \Rightarrow a_{1}=5 \lambda+2 \Rightarrow a_{1}=2 \Rightarrow 5 a_{2}+\ldots .5^{n-1} a_{n-1}=25 a_{2}+5 a_{3}+\ldots .5^{n-2} a_{n-2}=5 \Rightarrow a_{2}=5 \lambda \Rightarrow a_{2}=0 a_{3}+5 a_{4}+\ldots \ldots \ldots+5^{n-3} a_{n-3}=1 a_{3}=1 \Rightarrow a_{4}+5 a_{5}+\ldots .+5^{n-4} a_{n-3}=0 a_{4}=a_{5}=\ldots . a_{n}=0 Hence P(n)=x^{3}+2 x+1 P(3)=34 ## Subscribe to Cheenta at Youtube Categories ## Digits Problem | PRMO – 2018 | Question 19 Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19. ## Digits Problem – PRMO 2018, Question 19 Let N=6+66+666+\ldots \ldots+666 \ldots .66, where there are hundred 6 ‘s in the last term in the sum. How many times does the digit 7 occur in the number N ? • 30 • 33 • 36 • 39 • 42 ### Key Concepts Number theorm Digits Problem integer ## Check the Answer Answer:33 PRMO-2018, Problem 19 Pre College Mathematics ## Try with Hints Given that \mathrm{N}=6+66+666+……. \underbrace{6666 …..66}_{100 \text { times }} If you notice then we can see there are so many large terms. but we have to find out the sum of the digits. but since the number of digits are large so we can not calculate so eassily . we have to find out a symmetry or arrange the number so that we can use any formula taht we can calculate so eassily. if we multiply $\frac{6}{9}$ then it becomes =\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}] Can you now finish the problem ………. \mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}] =\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+…….+\left(10^{100}-1\right)\right] =\frac{6}{9}\left[\left(10+10^{2}+…..+10^{100}\right)-100\right] Can you finish the problem…….. =\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right] =\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60 =\frac{200}{27}\left(10^{99}-1\right)-60 =\frac{200}{27}\underbrace{(999….99)}_{99 \text{times}}-60 =\frac{1}{3}\underbrace{(222…..200)}_{99 \mathrm{times}}-60 =\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }} =\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340 \Rightarrow 7 comes 33 times ## Subscribe to Cheenta at Youtube Categories ## Chocolates Problem | PRMO – 2018 | Problem No. – 28 Try this beautiful Problem on Combinatorics from integer based on chocolates from PRMO -2018 ## Chocolates Problem – PRMO 2018- Problem 28 Let N be the number of ways of distributing 8 chocolates of different brands among 3 children such that each child gets at least one chocolate, and no two children get the same number of chocolates. Find the sum of the digits of \mathrm{N}. , • $28$ • $90$ • $24$ • $16$ • $27$ ### Key Concepts Combination Combinatorics Probability ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-28 #### Check the answer here, but try the problem first $24$ ## Try with Hints #### First Hint we have to distribute $8$ chocolates among $3$ childrens and the condition is Eight chocolets will be different brands that each child gets at least one chocolate, and no two children get the same number of chocolates. Therefore thr chocolates distributions will be two cases as shown below….. Now can you finish the problem? #### Second Hint case 1:(5,2,1) Out of $8$ chocolates one of the boys can get $5$ chocolates .So $5$ chocolates can be choosen from $8$ chocolates in $8 \choose 5$ ways. Therefore remaining chocolates are $3$ . Out of $3$ chocolates another one of the boys can get $2$ chocolates .So $2$ chocolates can be choosen from $3$ chocolates in $3 \choose 2$ ways. Therefore remaining chocolates are $1$ . Out of $1$ chocolates another one of the boys can get $1$ chocolates .So $1$ chocolates can be choosen from $1$ chocolates in $1 \choose 1$ ways. Therefore number of ways for first case will be $8 \choose 5$ $\times$ $3 \choose 2$ $\times$ $1 \choose 1$$\times$ 3!=\frac{8}{2!.5!.1!}$$\times 3$

Case 2:$(4,3,1)$

Out of $8$ chocolates one of the boys can get $4$ chocolates .So $4$ chocolates can be choosen from $8$ chocolates in $8 \choose 4$ ways.

Therefore remaining chocolates are $4$ . Out of $4$ chocolates another one of the boys can get $3$ chocolates .So $3$ chocolates can be choosen from $4$ chocolates in $4 \choose 3$ ways.

Therefore remaining chocolates are $1$ . Out of $1$ chocolates another one of the boys can get $1$ chocolates .So $1$ chocolates can be choosen from $1$ chocolates in $1 \choose 1$ ways.

Categories

## Trigonometry | PRMO-2018 | Problem No-14

Try this beautiful Problem on Trigonometry from PRMO -2018

## Trigonometry Problem – PRMO 2018- Problem 14

If $x=\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots . \cos 89^{\circ}$ and $y=\cos 2^{\circ} \cos 6^{\circ} \cos 10^{\circ} \ldots \ldots \cos 86^{\circ},$ then what is the integer nearest to $\frac{2}{7} \log _{2}(\mathrm{y} / \mathrm{x}) ?$

,

• $28$
• $19$
• $24$
• $16$
• $27$

Trigonometry

Logarithm

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

Prmo-2018, Problem-14

#### Check the answer here, but try the problem first

$19$

## Try with Hints

#### First Hint

Given that $x=\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \cdots \cos 89^{\circ}$

$\Rightarrow x=\cos 1^{\circ}\cos 89^{\circ}\cos 2^{\circ}\cos 88^{\circ}……..\cos 45^{\circ}$

$\Rightarrow x=\cos 1^{\circ}\sin 1^{\circ}\cos 2^{\circ}\cos 2^{\circ}……..\cos 45^{\circ}$ ( as cos(90-x)=sin x)

$\Rightarrow x=\frac{1}{2^{44}} (2 \cos 1^{\circ} \sin 1^{\circ}) (2\cos 2^{\circ}\sin 2^{\circ})……..\cos 45^{\circ}$

$\Rightarrow x =\frac{ \cos 45^{\circ} \cdot \sin 2^{\circ} \cdot \sin 4^{\circ} \cdots \sin 88}{ 2^{44}}$ (as sin 2x= 2 sin x cos x)

Now can you finish the problem?

#### Second Hint

Therefore we can say that
$x=\frac{1}{2^{1 / 2}} \times \frac{\sin 4^{\circ} \cdot \sin 8^{\circ} \cdots \sin 88^{\circ}}{2^{44} \times 2^{22}}$
$=\frac{\sin \left(90-86^{\circ}\right) \cdot \sin \left(90-84^{\circ}\right) \cdots \sin (90-2)}{2^{66} \cdot 2^{1 / 2}}$
$=\frac{\cos 2^{\circ} \cdot \cos 6^{\circ} \cdot \cos 86^{\circ}}{2^{133 / 2}}$
$x=\frac{y}{2^{133 / 2}}$

Can you finish the problem…?

#### Third Hint

$\frac{y}{x}=2^{133 / 2}$
$\frac{2}{7} \log \left(\frac{y}{x}\right)=\frac{2}{7} \times \log _{2}(2)^{133 / 2}$
$=\frac{2}{7} \times \frac{133}{2}$
$=19$

Categories

## Colour Problem | PRMO-2018 | Problem No-27

Try this beautiful Combinatorics Problem based on colour from integer from Prmo-2018.

## Colour Problem- PRMO 2018- Problem 27

What is the number of ways in which one can colour the square of a $4 \times 4$ chessboard with colours red and blue such that each row as well as each column has exactly two red squares and blue
squares?

,

• $28$
• $90$
• $32$
• $16$
• $27$

Chessboard

Combinatorics

Probability

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

Prmo-2018, Problem-27

#### Check the answer here, but try the problem first

$90$

## Try with Hints

#### First Hint

First row can be filled by ${ }^{4} \mathrm{C}_{2}$ ways $=6$ ways.
Case-I

Second row is filled same as first row
$\Rightarrow$
here second row is filled by one way
$3^{\text {rd }}$ row is filled by one way
$4^{\text {th }}$ row is filled by one way

Total ways in Case-I equals to ${ }^{4} \mathrm{C}_{1} \times 1 \times 1 \times 1=6$ ways

now we want to expand the expression and simplify it…………..

#### Second Hint

Case-II $\quad$ Exactly $1$ R & $1$ B is interchanged in second row in comparision to $1^{\text {st }}$ row
$\Rightarrow$
here second row is filled by $2 \times 2$ way
$3^{r d}$ row is filled by two ways
$4^{\text {th }}$ row is filled by one way
$\Rightarrow$
Total ways in Case-II equals to ${ }^{4} \mathrm{C}_{1} \times 2 \times 2 \times 2 \times 1=48$ ways

#### Third Hint

Case-III $\quad$ Both $\mathrm{R}$ and $\mathrm{B}$ is replaces by other in second row as compared to $1^{\text {st }}$ row
$\Rightarrow$
here second row is filled by 1 way
$3^{r d}$ row is filled by $4 \choose 2$ ways

$\Rightarrow \quad$ Total ways in $3^{\text {th }}$ Case equals to ${ }^{4} \mathrm{C}_{2} \times 1 \times 6 \times 1=36$ ways
$\Rightarrow \quad$ Total ways of all cases equals to 90 ways