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

But try the problem first…

Answer:$55$

Source

PRMO-2018, Problem 16

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

$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……..

Final Step

Therefore $S =(7+10+13+16+19)$-$(4-3-2-1)$ =$55$

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

But try the problem first…

Answer:$6$

Source

PRMO-2018, Problem 22

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

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……..

Final Step

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 $6$

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

But try the problem first…

Answer:$34$

Source

PRMO-2018, Problem 30

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

$a_{0}+a_{1}+a_{2}+\ldots \ldots+a_{n}=4$
$\Rightarrow a_{i} \leq 4$
$a_{0}+5 a_{1}+5^{2} a_{2}+\ldots+a 5^{n} a_{n}=136$
$\Rightarrow a_{0}=1+5 \lambda \Rightarrow a_{0}=1$

Can you finish the problem……..

Final Step

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

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

But try the problem first…

Answer:$33$

Source

PRMO-2018, Problem 19

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

$\mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
$=\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+…….+\left(10^{100}-1\right)\right]$
$=\frac{6}{9}\left[\left(10+10^{2}+…..+10^{100}\right)-100\right]$

Can you finish the problem……..

Final Step

$=\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right]$
$=\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60$
$=\frac{200}{27}\left(10^{99}-1\right)-60$
$=\frac{200}{27}\underbrace{(999….99)}_{99 \text{times}}-60$
$=\frac{1}{3}\underbrace{(222…..200)}_{99 \mathrm{times}}-60$

$=\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }}$ $=\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340$
$\Rightarrow 7$ comes 33 times

Categories

## Positive Integer | PRMO-2017 | Question 1

Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.

## Positive Integer – PRMO 2017, Question 1

How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by $3 ?$

• $9$
• $7$
• $28$

### Key Concepts

Algebra

Equation

multiplication

But try the problem first…

Answer:$28$

Source

PRMO-2017, Problem 1

Pre College Mathematics

## Try with Hints

First hint

Let $n$ be the positive integer less than 1000 and $s$ be the sum of its digits, then $3 \mid n$ and $7 \mid s$
$3|n \Rightarrow 3| s$
therefore$21| s$

Can you now finish the problem ……….

Second Hint

Also $n<1000 \Rightarrow s \leq 27$
therefore $\mathrm{s}=21$
Clearly, n must be a 3 digit number Let $x_{1}, x_{2}, x_{3}$ be the digits, then $x_{1}+x_{2}+x_{3}=21$
where $1 \leq x_{1} \leq 9,0 \leq x_{2}, x_{3} \leq 9$
$\Rightarrow x_{2}+x_{3}=21-x_{1} \leq 18$
$\Rightarrow x_{1} \geq 3$

Can you finish the problem……..

Final Step

For $x_{1}=3,4, \ldots ., 9,$ the equation (1) has $1,2,3, \ldots ., 7$ solutions
therefore total possible solution of equation (1)

=$1+2+\ldots+7=\frac{7 \times 8}{2}=28$

Categories

## Pen & Note Books Problem| PRMO-2017 | Question 8

Try this beautiful Pen & Note Books Problem from Algebra, from PRMO 2017.

## Pen & Note Books – PRMO 2017, Question 8

A pen costs $Rs 11$ and a notebook costs $Rs. 13 .$ Find the number of ways in which a person can spend exactly Rs.1000 to buy pens and notebooks.

• $9$
• $7$
• $11$

### Key Concepts

Algebra

Equation

multiplication

But try the problem first…

Answer:$7$

Source

PRMO-2017, Problem 8

Pre College Mathematics

## Try with Hints

First hint

Given A pen costs Rs.$11$ and a note book costs Rs.$13$

$11 x+13 y=1000$………………….(1)

Can you now finish the problem ……….

Second Hint

Now $11 x+13 y=1000$
$\Rightarrow 11 x=1000-13 y=(1001-11 y)-(2 y+1)$

=$11(91-y)-(2 y+1)$
$\Rightarrow 11 | 2 y+1$
Let $2 y+1=11(2 k-1), k \in I^{+}$
$\Rightarrow y=11 \mathrm{k}-6$
therefore $11 x=11(97-11 k)-11(2 k-1)$
$\Rightarrow x=98-13 k$
But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$

Can you finish the problem……..

Final Step

But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$
therefore for each $\mathrm{k} \in{1,2, \ldots ., 7},$ we get a unique pair $(\mathrm{x}, \mathrm{y})=(98-13 \mathrm{k}, 11 \mathrm{k}-6)$ satisfying equation
Hence 7 ways are possible.

Categories

## Rectangle Problem | Geometry | PRMO-2017 | Question 13

Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.

## Rectangle Problem – Geometry – PRMO 2017, Question 13

In a rectangle $A B C D, E$ is the midpoint of $A B ; F$ is a point on $A C$ such that $B F$ is perpendicular to $A C$; and FE perpendicular to BD. Suppose $\mathrm{BC}=8 \sqrt{3}$. Find AB.

• $9$
• $24$
• $11$

### Key Concepts

Geometry

Triangle

Trigonometry

But try the problem first…

Answer:$24$

Source

PRMO-2017, Problem 13

Pre College Mathematics

## Try with Hints

First hint

We have to find out the value of $AB$. Join $BD$. $BF$ is perpendicular on $AC$.

Let $\angle \mathrm{BAC}=\theta$
since $\mathrm{E}$ is mid point of hypotenous $\mathrm{AB}$ of right $\Delta \mathrm{AFB}$, therefore $A E=F E=B E$

Can you now finish the problem ……….

Second Hint

Therefore

Therefore$\angle E F A=\angle F A E=\theta$
and $\angle \mathrm{FEB}=\angle \mathrm{EAF}+\angle \mathrm{EFA}=2 \theta$
$\Rightarrow \angle E B D=90^{\circ}-\angle B E F=90^{\circ}-2 \theta$
But $\angle \mathrm{FAE}=\angle \mathrm{CAB}=\angle \mathrm{DBA}$
Therefore $\theta=90^{\circ}-2 \theta \Rightarrow \theta=30^{\circ}$
Therefore in $\Delta \mathrm{ABC}, \tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}$

Can you finish the problem……..

Final Step

Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$

Categories

## Distance travelled | PRMO II 2019 | Question 26

Try this beautiful problem from the Pre-RMO II, 2019, Question 26, based on Distance travelled.

## Distance travelled – Problem 26

A friction-less board has the shape of an equilateral triangle of side length 1 meter with bouncing walls along the sides. A tiny super bouncy ball is fired from vertex A towards the side BC. The ball bounces off the walls of the board nine times before it hits a vertex for the first time. The bounces are such that the angle of incidence equals the angle of reflection. The distance travelled by the ball in meters is of the form $\sqrt{N}$, where N is an integer

• is 107
• is 31
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

But try the problem first…

Source

PRMO II, 2019, Question 26

Higher Algebra by Hall and Knight

## Try with Hints

First hint

x= length of line segment

and by cosine law on triangle of side x, 1, 5 and 120 (in degrees) as one angle gives

$x^2=5^2+1^2-2 \times 5 \times 1 cos 120^\circ$

$=25+1+5$

Second Hint

or, x=$\sqrt{31}$

or, N=31

Final Step

Folding the triangle continuously each time of reflection creates the above diagram. 9 points of reflection can be seen in the diagram. Thus root (N) is the length of line which is root (31). Thus N=31 is the answer.

Categories

## Chords in a Circle | PRMO-2017 | Question 26

Try this beautiful Problem based on Chords in a Circle, Geometry, from PRMO 2017.

## Chords in a Circle – PRMO 2017, Question 26

Let $A B$ and $C D$ be two parallel chords in a circle with radius 5 such that the centre $O$ lies between these chords. Suppose $A B=6, C D=8 .$ Suppose further that the area of the part of the circle lying between the chords $A B$ and $C D$ is $(m \pi+n) / k,$ where $m, n, k$ are positive integers with gcd$(m, n, k)=1$ . What is the value of $m+n+k ?$

• $9$
• $75$
• $11$

### Key Concepts

Geometry

Triangle

Circle

But try the problem first…

Answer:$75$

Source

PRMO-2017, Problem 26

Pre College Mathematics

## Try with Hints

First hint

Draw OE $\perp A B$ and $O F \perp C D$

Clearly $\mathrm{EB}=\frac{\mathrm{AB}}{2}=3, \mathrm{FD}=\frac{\mathrm{CU}}{2}=4$

$\mathrm{OE}=\sqrt{5^{2}-3^{2}}=4$ and $\mathrm{OF}=\sqrt{5^{2}-4^{2}}=3$

Therefore $\Delta O E B \sim \Delta D F O$

Can you now finish the problem ……….

Second Hint

Let $\angle \mathrm{EOB}=\angle \mathrm{ODF}=\theta,$ then

$\angle B O D=\angle A O C=180^{\circ}-\left(\theta+90^{\circ}-\theta\right)=90^{\circ}$

Now area of portion between the chords

= $2 \times$ (area of minor sector BOD)+2 \times ar$(\triangle AOB)$
$=2 \times \frac{\pi \times 5^{2}}{4}+2 \times \frac{1}{2} \times 6 \times 4=\frac{25 \pi}{2}+24=\frac{25 \pi+48}{2}$

Therefore $m=25, n=48$ and $k=2$

Can you finish the problem……..

Final Step

Therefore $m+n+k=75$

Categories

## Shortest Distance | PRMO II 2019 | Question 27

Try this beautiful problem from the Pre-RMO II, 2019, Question 27 based on Shortest Distance.

## Shortest Distance – Pre-RMO II, Problem 27

A conical glass is in the form of a right circular cone. The slant height is 21 and the radius of the top rim of the glass is 14. An ant at the mid point of a slant line on the outside wall of the glass sees a honey drop diametrically opposite to it on the inside wall of the glass. If d the shortest distance it should crawl to reach the honey drop, what is the integer part of d?

• is 107
• is 36
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

But try the problem first…

Source

PRMO II, 2019, Question 27

Higher Algebra by Hall and Knight

## Try with Hints

First hint

Rotate $\Delta$OAP by 120$^\circ$ in anticlockwise then A will be at B, P will be at P’

Second Hint

or, $\Delta$OAP is congruent to $\Delta$OBP’

or, PB+PA=P’B+PB $\geq$ P’P

Minimum PB+PA=P’P equality when P on the angle bisector of $\angle$AOB

or, P’P=2(21)sin60$^\circ$=21$\sqrt{3}$

Final Step

[min(PB+PA)]=[21$\sqrt{3}$]=36 (Answer)