Category: Algebra

Value of Sum | PRMO – 2018 | Question 16

Post author By KOUSHIK SOM
Post date October 16, 2020
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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

But try the problem first…

Answer:$55$

Source

Suggested Reading

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 $

Digits Problem | PRMO – 2018 | Question 19

Post author By KOUSHIK SOM
Post date October 6, 2020
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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

But try the problem first…

Answer:$33$

Source

Suggested Reading

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

Problem on Curve | AMC 10A, 2018 | Problem 21

Post author By KOUSHIK SOM
Post date September 29, 2020
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Try this beautiful Problem on Algebra based on Problem on Curve from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Curve- AMC 10A, 2018- Problem 21

Which of the following describes the set of values of $a$ for which the curves $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ in the real $x y$ -plane intersect at
exactly 3 points?

$a=\frac{1}{4}$
$\frac{1}{4}<a<\frac{1}{2}$
$a>\frac{1}{4}$
$a=\frac{1}{2}$
$a>\frac{1}{2}$

Key Concepts

Algebra

greatest integer

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2018 Problem-14

Check the answer here, but try the problem first

$a>\frac{1}{2}$

Try with Hints

First Hint

We have to find out the value of $a$

Given that $y=x^{2}-a$ . now if we Substitute this value in $x^{2}+y^{2}=a^{2}$ we will get a quadratic equation of $x$ and $a$. if you solve this equation you will get the value of $a$

Now can you finish the problem?

Second Hint

After substituting we will get $x^{2}+\left(x^{2}-a\right)^{2}$=$a^{2} \Longrightarrow x^{2}+x^{4}-2 a x^{2}=0 \Longrightarrow x^{2}\left(x^{2}-(2 a-1)\right)=0$

therefore we can say that either $x^2=0\Rightarrow x=0$ or $x^2-(2a-1)=0$

$\Rightarrow x=\pm \sqrt {2a-1}$. Therefore

Now Can you finish the Problem?

Third Hint

Therefore $\sqrt {2a-1} > 0$

$\Rightarrow a>\frac{1}{2}$

Finding Greatest Integer | AMC 10A, 2018 | Problem No 14

Post author By KOUSHIK SOM
Post date September 28, 2020
1 Comment on Finding Greatest Integer | AMC 10A, 2018 | Problem No 14

Try this beautiful Problem on Algebra based on finding greatest integer from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Finding Greatest Integer – AMC-10A, 2018- Problem 14

What is the greatest integer less than or equal to $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

$80$
$81$
$96$
$97$
$625$

Key Concepts

Algebra

greatest integer

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2018 Problem-14

Check the answer here, but try the problem first

$80$

Try with Hints

First Hint

The given expression is $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

We have to find out the greatest integer which is less than or equal to the given expression .

Let us assaume that $x=3^{96}$ and $y=2^{96}$

Therefore the given expression becoms $\frac{81 x+16 y}{x+y}$

Now can you finish the problem?

Second Hint

Now $\frac{81 x+16 y}{x+y}$

=$\frac{16 x+16 y}{x+y}+\frac{65 x}{x+y}$

$=16+\frac{65 x}{x+y}$

Now if we look very carefully we see that $\frac{65 x}{x+y}<\frac{65 x}{x}=65$

Therefore $16+\frac{65 x}{x+y}<16+65=81$

Now Can you finish the Problem?

Third Hint

Therefore less than $81$ , the answer will be $80$

Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Post author By KOUSHIK SOM
Post date September 19, 2020
2 Comments on Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Sum of Sides of Triangle from Triangle from Prmo-2018, Problem 17.

Sum of Sides of Triangle – PRMO, 2018- Problem 17

Triangles $\mathrm{ABC}$ and $\mathrm{DEF}$ are such that $\angle \mathrm{A}=\angle \mathrm{D}, \mathrm{AB}=\mathrm{DE}=17, \mathrm{BC}=\mathrm{EF}=10$ and $\mathrm{AC}-\mathrm{DF}=12$
What is $\mathrm{AC}+$ DF ?

$28$
$30$
$21$
$26$
$26$

Key Concepts

Geometry

Triangle

Suggested Book | Source | Answer

Source of the problem

Prmo-2018, Problem-17

Check the answer here, but try the problem first

$30$

Try with Hints

First Hint

In $\triangle ABC$ & $\triangle DEF$, given that $\angle A$=$\angle D$ & $AB=DE$=$17$ . $BC = EF = 10 $ and $AC – DF = 12.$ we have to find out $AC+DF$.

According to the question Let us assume that the point A coincides with D, B coincides with E. Now if we draw a circle with radius 10 and E(B) as center ….

Can you finish the problem?

Second Hint

Let M be the foot of perpendicular from B(E) to CF. So BM = 8. Hence $\mathrm{AM}=\sqrt{17^{2}-8^{2}}=\sqrt{(25)(9)}=15$

Third Hint

Hence $AF = 15 – 6 = 9$ & $AC = 15 + 6 = 21$

So $AC + DF = 30$

Linear Equation Problem | AMC 10A, 2015 | Problem No.16

Post author By KOUSHIK SOM
Post date August 21, 2020
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Try this beautiful Problem on Algebra from the Linear equation from AMC 10 A, 2015.

Linear Equation Problem – AMC-10A, 2015- Problem 16

If $y+4=(x-2)^{2}, x+4=(y-2)^{2}$, and $x \neq y$, what is the value of $x^{2}+y^{2} ?$

$11$
$12$
$15$
$14$
$6$

Key Concepts

Algebra

Equation

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2015 Problem-16

Check the answer here, but try the problem first

$15$

Try with Hints

First Hint

Given that $y+4=(x-2)^{2}, x+4=(y-2)^{2}$ . we have to find out $x^{2}+y^{2} ?$. Now add two equations $x^{2}+y^{2}-4 x-4 y+8=x+y+8$

$\Rightarrow x^{2}+y^{2}=5(x+y)$

Can you find out the value $x+y$?

Second Hint

We can also subtract the two equations to yield the equation
$x^{2}-y^{2}-4 x+4 y=y-x$

$\Rightarrow x^{2}-y^{2}=(x+y)(x-y)=3 x-3 y=3(x-y)$

Therefore $(x+y)(x-y)=3 x-3 y=3(x-y)$

$\frac{(x+y)(x-y)}{(x-y)}=\frac{3(x-y)}{(x-y)}$ [ as$ x \neq y$]

$\Rightarrow (x+y)=3$

Third Hint

Therefore $x^2+y^2=5(x+y)=5 \times 3=15$

Least Possible Value Problem | AMC-10A, 2019 | Quesstion19

Post author By KOUSHIK SOM
Post date August 2, 2020
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Try this beautiful problem from Algebra based on Least Possible Value.

Least Possible Value – AMC-10A, 2019- Problem 19

What is the least possible value of $((x+1)(x+2)(x+3)(x+4)+2019)$

where (x) is a real number?

$(2024)$
$(2018)$
$(2020)$

Key Concepts

Algebra

quadratic equation

least value

Check the Answer

But try the problem first…

Answer: $(2018)$

Source

Suggested Reading

AMC-10A (2019) Problem 19

Pre College Mathematics

Try with Hints

First hint

To find out the least positive value of $(x+1)(x+2)(x+3)(x+4)+2019$, at first we have to expand the expression .$((x+1)(x+2)(x+3)(x+4)+2019)$ $\Rightarrow (x+1)(x+4)(x+2)(x+3)+2019=(x^2+5x+4)(x^2+5x+6)+2019)$

Let us take $((x^2+5x+5=m))$

then the above expression becomes $((m-1)(m+1)+2019)$ $\Rightarrow m^2-1+2019$ $\Rightarrow m^2+2018$

Can you now finish the problem ……….

Second Hint

Clearly in $(m^2+2018)…….(m^2)$ is positive ( squares of any number is non-negative) and least value is 0

can you finish the problem……..

Final Step

Therefore minimum value of $m^2+2108$ is $2018$ since $m^2 \geq 0$ for all m belongs to real .

Nearest value | PRMO 2018 | Question 14

Post author By Shirsendu Roy
Post date August 1, 2020
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Try this beautiful problem from the PRMO, 2018 based on Nearest value.

Nearest Value – PRMO 2018

If x=cos1cos2cos3…..cos89 and y=cos2cos6cos10….cos86, then what is the integer nearest to $\frac{2}{7}log_2{\frac{y}{x}}$?

is 107
is 19
is 840
cannot be determined from the given information

Key Concepts

Algebra

Numbers

Multiples

Check the Answer

But try the problem first…

Answer: is 19.

Source

Suggested Reading

PRMO, 2018, Question 14

Higher Algebra by Hall and Knight

Try with Hints

First hint

$\frac{y}{x}$=$\frac{cos2cos6cos10…..cos86}{cos1cos2cos3….cos89}$

=$2^{44}\times\sqrt{2}\frac{cos2cos6cos10…cos86}{sin2sin4…sin88}$

[ since cos$\theta$=sin(90-$\theta$) from cos 46 upto cos 89 and 2sin$\theta$cos$\theta$=sin2$\theta$]

Second Hint

=$\frac{2^{\frac{89}{2}}sin4sin8sin12…sin88}{sin2sin4sin6…sin88}$

[ since sin$\theta$=cos(90-$\theta$)]

=$\frac{2^{\frac{89}{2}}}{cos4cos8cos12..cos88}$

[ since cos$\theta$=sin(90-$\theta$)]

Final Step

=$\frac{2^\frac{89}{2}}{\frac{1}{2}^{22}}$

[since $cos4cos8cos12…cos88$

$=(cos4cos56cos64)(cos8cos52cos68)(cos12cos48cos72)(cos16cos44cos76)(cos20cos40cos80)(cos24cos36cos84)(cos28cos32cos88)cos60$

$=(1/2)^{15}(cos12cos24cos36cos48cos60cos72cos84)$

$=(1/2)^{16}(cos12cos48cos72)(cos24cos36cos84)$

$=(1/2)^{20}(cos36cos72)$

$=(1/2)^{20}(cos36sin18)$

$=(1/2)^{22}(4sin18cos18cos36/cos18)$

$=(1/2)^{22}(sin72/cos18)$

$=(1/2)^{22}$]

=$2^\frac{133}{2}$

$\frac{2}{7}log_2{\frac{y}{x}}$=$\frac{2}{7} \times \frac{133}{2}$=19.

Sequence and permutations | AIME II, 2015 | Question 10

Post author By Shirsendu Roy
Post date July 19, 2020
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Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

Sequence and permutations – AIME II, 2015

Call a permutation $a_1,a_2,….,a_n$ of the integers 1,2,…,n quasi increasing if $a_k \leq a_{k+1} +2$ for each $1 \leq k \leq n-1$, find the number of quasi increasing permutations of the integers 1,2,….,7.

is 107
is 486
is 840
cannot be determined from the given information

Key Concepts

Sequence

Permutations

Integers

Check the Answer

But try the problem first…

Answer: is 486.

Source

Suggested Reading

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

Try with Hints

First hint

While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Second Hint

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 $\times$ number of permutations of (n-1) elements

or, number of permutations for n elements=$3^{2}$ number of permutations of (n-2) elements

……

or, number of permutations for n elements=$3^{n-2}$ number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 $\times$ $3^{n-2}$

forming recurrence relation as the number of permutations =2 $\times$ $3^{n-2}$

for n=3 all six permutations taken and go up 18, 54, 162, 486

Final Step

for n=7, here $2 \times 3^{5} =486.$

Header text

sds

Numbers of positive integers | AIME I, 2012 | Question 1

Post author By Shirsendu Roy
Post date July 17, 2020
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Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.

Numbers of positive integers – AIME 2012

Find the number of positive integers with three not necessarily distinct digits, $abc$, with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$.

is 107
is 40
is 840
cannot be determined from the given information

Key Concepts

Integers

Number Theory

Algebra

Check the Answer

But try the problem first…

Answer: is 40.

Source

Suggested Reading

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

Try with Hints

First hint

Here a number divisible by 4 if a units with tens place digit is divisible by 4

Second Hint

Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b

[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]

and case II 2(mod4) with a pair of a and c for every b

Then combining both cases we get for every b gives a pair of a s and a pair of c s

Final Step

So for 10 b’s with 2 a’s and 2 c’s for every b gives $10 \times 2 \times 2$

Then number of ways $10 \times 2 \times 2$ = 40 ways.

Value of Sum – PRMO 2018, Question 16

Key Concepts

Check the Answer

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Digits Problem – PRMO 2018, Question 19

Key Concepts

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Curve- AMC 10A, 2018- Problem 21

Key Concepts

Suggested Book | Source | Answer

Suggested Reading

Source of the problem

Check the answer here, but try the problem first

Try with Hints

First Hint

Second Hint

Third Hint

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Finding Greatest Integer – AMC-10A, 2018- Problem 14

Key Concepts

Suggested Book | Source | Answer

Suggested Reading

Source of the problem

Check the answer here, but try the problem first

Try with Hints

First Hint

Second Hint

Third Hint

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Subscribe to Cheenta at Youtube

Sum of Sides of Triangle – PRMO, 2018- Problem 17

Key Concepts

Suggested Book | Source | Answer

Suggested Reading

Source of the problem

Check the answer here, but try the problem first

Try with Hints

First Hint

Second Hint

Third Hint

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Linear Equation Problem – AMC-10A, 2015- Problem 16

Key Concepts

Suggested Book | Source | Answer

Suggested Reading

Source of the problem

Check the answer here, but try the problem first

Try with Hints

First Hint

Second Hint

Third Hint

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Least Possible Value – AMC-10A, 2019- Problem 19

Key Concepts

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Nearest Value – PRMO 2018

Key Concepts

Check the Answer

Try with Hints

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