Categories
Math Olympiad PRMO

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 $\mathrm{N} / 100$ ?

,

  • \(15\)
  • \(22\)
  • \(27\)
  • \(32\)
  • \(37\)

Key Concepts


Trigonometry

Triangle

Integer

Suggested Book | Source | Answer


Suggested Reading

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 \
$

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Categories
Algebra Math Olympiad PRMO USA Math Olympiad

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

Check the Answer


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 $



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Categories
Combinatorics Math Olympiad PRMO

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 ${ }^{61} \mathrm{C}{60}+{ }^{60} \mathrm{C}{60}=61+1=62$

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Categories
Math Olympiad PRMO

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

Measure of Angle

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 $\quad \theta=30^{\circ}$

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Categories
Math Olympiad PRMO USA Math Olympiad

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 $6$

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Categories
Math Olympiad PRMO USA Math Olympiad

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 $P(n)=x^{3}+2 x+1$
$P(3)=34$

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Categories
Algebra Math Olympiad PRMO USA Math Olympiad

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



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Categories
Combinatorics Math Olympiad PRMO

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.

Therefore number of ways for first case will be \( 8 \choose 4\) \( \times\) \( 4 \choose 3\) \( \times\) \( 1 \choose 1\)\(\times\) $3!$=$\frac{8}{4!.3!.1!}$$\times 3$

Can you finish the problem…?

Third Hint


Therefore require number of ways =$\frac{8}{2!.5!.1!}$$\times 3$+$\frac{8}{4!.3!.1!}$$\times 3$=$24$

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Combinatorics Math Olympiad PRMO

Trigonometry | PRMO-2018 | Problem No-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 $\frac{2}{7} \log _{2}(\mathrm{y} / \mathrm{x}) ?$

,

  • \(28\)
  • \(19\)
  • \(24\)
  • \(16\)
  • \(27\)

Key Concepts


Trigonometry

Logarithm

Suggested Book | Source | Answer


Suggested Reading

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}$
$=\frac{2}{7} \times \frac{133}{2}$
$=19$

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Categories
Combinatorics Math Olympiad PRMO

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?

,

  • \(28\)
  • \(90\)
  • \(32\)
  • \(16\)
  • \(27\)

Key Concepts


Chessboard

Combinatorics

Probability

Suggested Book | Source | Answer


Suggested Reading

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


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