Categories

## Largest Possible Value | PRMO-2019 | Problem 17

Try this beautiful problem from PRMO, 2019 based on Largest Possible Value.

## Largest Possible Value | PRMO | Problem-17

Let a, b, c be distinct positive integers such that $b + c – a$,$c + a – b$ and $a + b – c$ are all perfect squares.
What is the largest possible value of $a + b + c$ smaller than $100$?

• $20$
• $91$
• $13$

### Key Concepts

Number theory

Perfect square

Integer

Answer:$91$

PRMO-2019, Problem 17

Pre College Mathematics

## Try with Hints

Let $b + c – a = x^2$ … (i)
$c + a – b = y^2$ … (ii)
$a + b – c = z^2$ … (iii)

Now since $a$,$b$, $c$ are distinct positive integers,
Therefore, $x$, $y$, $z$ will also be positive integers,
$a + b + c = x^2 + y^2 + z^2$
Now, we need to find largest value of $a + b + c or x^2 + y^2 + z^2$ less than $100$
Now, to get a, b, c all integers $x$,$y$, $z$ all must be of same parity, i.e. either all three are even or all three
are odd.

Can you now finish the problem ……….

Let us maximize$x^2 + y^2 + z^2$, for both cases.
If $x$, $y$, $z$are all even.
Therefore,

$b + c – a = 8^2 = 64$
$c + a – b = 42 = 16$
$a + b – c = 22 = 4$
Which on solving, give$a = 10$,$b = 34$, $c = 40$ and $a + b + c = 84$
If x, y, z are all odd
$\Rightarrow b + c – a = 92 = 81$
$c + a – b = 32 = 9$
$a + b – c = 12 = 1$
Which on solving, give $a = 5$ ,$b = 41$, $c = 45$ and$a + b + c = 91$

Can you finish the problem……..

Therefore Maximum value of $a + b + c < 100 = 91$

Categories

## Smallest positive value | Algebra | PRMO-2019 | Problem 13

Try this beautiful problem from Algebra PRMO 2019 based on smallest positive value

## Smallest positive value| PRMO | Problem 13

Each of the numbers $x_1, x_2,……….x_{101}$ is $±1$. What is the smallest positive value of $\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j$ ?

• $24$
• $10$
• $34$

### Key Concepts

Algebra

Integer

sum

Answer:$10$

PRMO-2019, Problem 13

Pre College Mathematics

## Try with Hints

$S=\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j$ .

we have $(x_1+x_2+x_3+….+x_{101})^2={x_1}^2+{x_2}^2+…..+{x_{101}}^2+2S$

$\Rightarrow 2S$=$(\displaystyle\sum_{i=1}^{101} {x_i})^2-\displaystyle\sum_{j=1}^{101} {x_j}^2$

Can you now finish the problem ……….

Since we have $x_i=\pm 1$ so ${x_i}^2=1$

so $2S=(\displaystyle\sum_{i=1}^{101} {x_i})^2-{101}$

Since $\displaystyle\sum_{i=1}^{101} {x_i}$ will be an integer

so $(\displaystyle\sum_{i=1}^{101} {x_i})^2$ will be a perfect square .

For smalll positive $S$, $(\displaystyle\sum_{i=1}^{101} {x_i})^2$must be smallest perfect square greater than ${101}$

So $(\displaystyle\sum_{i=1}^{101} {x_i})^2={121}$

$\Rightarrow (\displaystyle\sum_{i=1}^{101} {x_i})$=${11}$ or ${-11}$

Can you finish the problem……..

We can verify that the desired sum can be achieved by putting $45$ $x_i$’s to be –1 and $56$ $x_i$’s to be $1$ So, $2S = 121 – 101 = 20$

$\Rightarrow s=10$

Categories

## Regular polygon | Combinatorics | PRMO-2019 | Problem 15

Try this beautiful problem from combinatorics PRMO 2019 based on Regular polygon

## Regular polygon| PRMO | Problem 15

In how many ways can a pair of parallel diagonals of a regular polygon of $10$ sides be selected

• $24$
• $45$
• $34$

### Key Concepts

Combinatorics

Regular polygon

geometry

Answer:$45$

PRMO-2019, Problem 15

Pre College Mathematics

## Try with Hints

The above diagram is a diagram of Regular Polygon .we have to draw the diagonals as shown in above.we joined the diagonals such that all the diagonals will be parallel

Can you now finish the problem ……….

If we joined the diagonals (shown in Fig. 1), i.e $(P_3 \to P_10)$,$(P_4\to P_9)$,$(P_5 \to P_8)$ then then we have 3 diagonals.so we have 5$4 \choose 2$ ways=$15$ ways.

If we joined the diagonals (shown in Fig.2), i.e $(P_1 \to P_3)$,$(P_10\to P_4)$,$(P_9\to P_5)$,$(P_8\to P_6)$then we have $4$diagonals.so we have 5$3 \choose 2$ ways=$30$ ways.

Therefore total numbers of ways that can a pair of parallel diagonals of a regular polygon of $10$ sides be selected is $15+30=45$

Categories

## Good numbers Problem | PRMO-2019 | Problem 12

Try this beautiful problem from PRMO, 2019 based on Good numbers.

## Good numbers Problem | PRMO | Problem-12

A natural number $k >$ is called good if there exist natural numbers
$a_1 < a_2 < ………. < a_k$

$\frac{1}{\sqrt a_1} +\frac{1}{\sqrt a_2}+………………. +\frac{1}{\sqrt a_k}=1$

Let $f(n)$ be the sum of the first $n$ good numbers, $n \geq 1$. Find the sum of all values of $n$ for which
$f(n + 5)/f(n)$ is an integer.

• $20$
• $18$
• $13$

### Key Concepts

Number theory

Good number

Integer

Answer:$18$

PRMO-2019, Problem 12

Pre College Mathematics

## Try with Hints

A number n is called a good number if It is a square free number.

Let $a_1 ={A_1}^2$,$a_2={A_2}^2$,………………$a_k={A_k}^2$
we have to check if it is possible for distinct natural number $A_1, A_2………….A_k$ to satisfy,
$\frac{1}{A_1}+\frac{1}{A_2}+………..+\frac{1}{A_k}=1$

Can you now finish the problem ……….

For $k = 2$; it is obvious that there do not exist distinct$A_1, A_2$, such that $\frac{1}{A_1}+\frac{1}{A_2}=1 \Rightarrow 2$ is not a good number

For $k = 3$; we have $\frac{1}{2} +\frac{1}{3}+\frac{1}{6}=1 \Rightarrow 3$ is a good number.

$\frac{1}{2}+\frac{1}{2}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ $\Rightarrow 4$ is a good number

Let $k$ wil be a good numbers for all $k \geq 3$

$f(n) = 3 + 4 +… n$ terms =$\frac{n(n + 5)}{2}$
$f(n + 5) =\frac{(n + 5)(n +10)}{2}$

$\frac{f(n+5}{f(n)}=\frac{n+10}{n}=1+\frac{10}{n}$

Can you finish the problem……..

Therefore the integer for n = $1$, $2$, $5$ and $10$. so sum=$1 + 2 + 5 + 10 = 18$.

Categories

## Quadratic equation | ISI-B.stat | Objective Problem 240

Try this beautiful problem based on Quadratic equation, useful for ISI B.Stat Entrance.

## Quadratic equation | ISI B.Stat Entrance | Problem 240

The equations $x^2 + x + a = 0$ and $x^2 + ax + 1 = 0$

• (a) cannot have a common real root for any value of a
• (b) have a common real root for exactly one value of a
• (c) have a common root for exactly two values of a
• (d) have a common root for exactly three values of a.

### Key Concepts

Algebra

Roots

TOMATO, Problem 240

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Let the equations have a common root $α$.Therefore $α$ must satisfy two given equations…….

Therefore,

Now, $α^2 + α + a = 0$……………….(1)
And, $α^2 + aα + 1 = 0$…………………..(2)

Can you find out the value of $a$?

Can you now finish the problem ……….

Therefore,

Using cross-multiplication betwwen (1) & (2) we will get…….

$\frac{α^2}{(1 – a^2)} =\frac{ α}{(a – 1)} = \frac{1}{(a – 1)}$
$\Rightarrow {α}^2 = \frac{(1 – a^2)}{(a – 1) }=- (a + 1)$ & $α=\frac{(a-1)}{(a-1)}=1$

Now ${α}^2=(α)^2$
$\Rightarrow -(a+1)=1$
$\Rightarrow a = -2$

Therefore (b) is the correct answer….

Categories

## Hexagon Problem | Geometry | AMC-10A, 2010 | Problem 19

Try this beautiful problem on area of Hexagon from Geometry.

## Hexagon Problem – AMC-10A, 2010- Problem 19

Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?

• $6$
• $\frac{2}{\sqrt5}$
• $8$
• $9$
• $9$

### Key Concepts

Geometry

Hexagon

Triangle

Answer: $6$

AMC-10A (2010) Problem 19

Pre College Mathematics

## Try with Hints

Given that area of $\triangle ACE$ is $70$% of the area of the hexagon.so at first we have to find out the area of the $\triangle ACE$.Clearly $\triangle ACE$ is an equilateral triangle.Now from the cosines law we can say that $AC^2=r^2+1^2-2r cos \frac{2\pi}{3}=r^2+r+1$.Therefore area of $\triangle ACE=\frac{\sqrt 3}{4}(r^2+r+1)$.Can you find out area of the Hexagon $ABCDEF$?

Can you now finish the problem ……….

Area Of The Hexagon $ABCDEF$ :

If we extend $BC$, $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$, $BC$ and $DE$ meet at $Y$, and $DE$ and $FA$ meet at $Z$,Now we can find out area of the hexagon $ABCDEF$ which is formed by taking equilateral triangle $XYZ$ of side length $r+2$. So if we remove three equilateral triangles, $ABX$, $CDY$ and $EFZ$, of side length $1$.

Therefore The area of $ABCDEF$ is $\frac{\sqrt 3}{4} (r+2)^2 – \frac{3\sqrt 3}{4}=\frac{\sqrt 3}{4} (r^2+4r+1)$

can you finish the problem……..

Given that area of $\triangle ACE$ is $70$% of the area of the hexagon.Therefore

$\frac{\sqrt 3}{4}(r^2+r+1)$=$\frac{7}{10}.\frac{\sqrt 3}{4}(r^2+4r+1)$

$\Rightarrow r^2-6r+1=0$.Now from Vieta’s Relation the sum of the possible value of $r$ is $6$

Categories

## Integer Problem | AMC 10A, 2020 | Problem 17

Try this beautiful problem from Number theory based on Integer.

## Integer Problem – AMC-10A, 2020- Problem 17

Define $P(x)=(x-1^2)(x-2^2)……(x-{100}^2)$

How many integers $n$ are there such that $P(n) \geq 0$?

• $4900$
• $4950$
• $5000$
• $5050$
• $5100$

### Key Concepts

Number system

Probability

divisibility

Answer: $5100$

AMC-10A (2020) Problem 17

Pre College Mathematics

## Try with Hints

Given $P(x)=(x-1^2)(x-2^2)……(x-{100}^2)$. at first we notice that $P(x)$ is a product of of $100$ terms…..now clearly $P(x)$ will be negetive ,for there to be an odd number of negetive factors ,n must be lie between an odd number squared and even number squared.

can you finish the problem……..

$P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2$, $98^2$ and $97^2 \ldots$ , $2^2$ and $1^2$

can you finish the problem……..

Therefore, $(100+99)(100-99)+((98+97)(98-97)+1)$+….

.+$((2+1)(2-1)+1)$=$(200+196+192+…..+4)$ =$4(1+2+…..+50)=4 \frac{50 \times 51}{2}=5100$

Categories

## Time & Work Problem | PRMO-2017 | Problem 3

Try this beautiful problem from PRMO, 2017 from Arithmetic based on Time & Work.

## Time & Work Problem | PRMO | Problem 3

A contractor has two teams of workers: team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins A after four days. Team A withdraws after two more days. For how many more days should team B work to complete the job?

• $24$
• $16$
• $22$
• $18$

### Key Concepts

Arithmetic

Unitary process

Work done

Answer:$16$

PRMO-2017, Problem 3

Pre College Mathematics

## Try with Hints

At first we have to find out A’s 1 days work and B’s 1 days work.next find out A and B both together 1 day’s work .

Can you now finish the problem ……….

Team A completes job in 12 days and Team B completes job in 36 days

1 day work of team A =$\frac{1}{12}$

1 day work of team B=$\frac{1}{36}$

1 day work of team A and team B (when they both work together $\frac{1}{12} +\frac{1}{36}$=$\frac{1}{9}$

Now according to question,
Let more number of days should team B works to complete the job be x days

$4 \times \frac{1}{12} +2 \times \frac{1}{9} + x \times \frac{1}{36}=1$

$\Rightarrow x=16$

Categories

## Problem on Area of Trapezoid | AMC-10A, 2002 | Problem 25

Try this beautiful problem on area of trapezoid from Geometry.

## Problem on Area of Trapezoid – AMC-10A, 2002- Problem 25

In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$. The area of $ABCD$ is

• $182$
• $195$
• $210$
• $234$
• $260$

### Key Concepts

Geometry

Trapezoid

Triangle

Answer: $210$

AMC-10A (2002) Problem 25

Pre College Mathematics

## Try with Hints

Given that $ABCD$ is a Trapezium with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$.we have to find out the area of the Trapezium.Normally the area of the trapezium is $\frac{1}{2} (AD +BC) \times$(height between CD & AB).but we don’t know the height.So another way if we extend $AD$ & $BC$ ,they will meet a point $E$.Now clearly Area of $ABCD$=Area of $\triangle ABE$ – Area of $\triangle EDC$.Can you find out the area of $\triangle EAB$ & Area of $\triangle EDC$?

Can you now finish the problem ……….

Now $AB||DC$ , Therefore $\triangle EDC \sim \triangle EAB$

$\Rightarrow \frac{ED}{EA}=\frac{EC}{EB}=\frac{DC}{AB}$

$\Rightarrow \frac{ED}{ED+DA}=\frac{EC}{EC+BC}=\frac{DC}{AB}$

$\Rightarrow \frac{ED}{ED+5}=\frac{EC}{EC+12}=\frac{39}{52}$

Now , $\frac{ED}{ED+5}=\frac{39}{52}$

$\Rightarrow ED=15$

And $\frac{EC}{EC+12}=\frac{39}{52}$

$\Rightarrow CE=36$

Therefore $BE$=$12+36=48$ and $AE=20$

Notice that in the $\triangle EDC$, ${ED}^2 +{EC}^2=(36)^2+(15)^2=(39)^2=(DC)^2$ $\Rightarrow \triangle EDC$ is a Right-angle Triangle

Therefore Area of $\triangle EDC=\frac{1}{2} \times 36 \times 15=270$

Similarly In the $\triangle EAB$, ${EA}^2 +{EB}^2=(48)^2+(20)^2=(52)^2=(AB)^2$ $\Rightarrow \triangle EAB$ is a Right-angle Triangle

Therefore Area of $\triangle EAB=\frac{1}{2} \times 48 \times 20=480$

Now can you find out the area of $ABCD$?

can you finish the problem……..

Therefore Area of $ABCD$=Area of $\triangle ABE$ – Area of $\triangle EDC$=$480-270=210$

Categories

## Quadratic equation Problem | AMC-10A, 2002 | Problem 12

Try this beautiful problem from Algebra based Quadratic equation.

## Quadratic equation Problem – AMC-10A, 2002- Problem 12

Both roots of the quadratic equation $x^2 – 63x + k = 0$ are prime numbers. The number of possible values of $k$ is

• $0$
• $1$
• $2$
• $4$
• more than $4$

### Key Concepts

Algebra

prime numbers

Answer: $1$

AMC-10A (2002) Problem 12

Pre College Mathematics

## Try with Hints

The given equation is $x^2 – 63x + k = 0$. Say that the roots are primes…

Comparing the equation with $ax^2 +bx+c=0$ we get $a=1 , b=-63 , c=k$.. Let $m_1$ & $m_2$ be the roots of the given equation…

using vieta’s Formula we may sat that…$m_1 + m_2 =-(- 63)=63$ and $m_1 m_2 = k$

can you finish the problem……..

Now the roots are prime. Sum of the two roots are $63$ and product is $k$

Therefore one root must be $2$ ,otherwise the sum would be even number

can you finish the problem……..

So other root will be $63-2$=$61$. Therefore product must be $m_1m_2=122$

Hence the answer is $1$