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

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## 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

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

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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

least value

Answer: $(2018)$

AMC-10A (2019) Problem 19

Pre College Mathematics

## Try with Hints

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 ……….

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……..

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

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## Nearest value | PRMO 2018 | Question 14

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

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.

Categories

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

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

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

## Try with Hints

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

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

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.

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## Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

## Arithmetic Sequence Problem – AIME 2012

The terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.

• is 107
• is 195
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Number Theory

Algebra

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

## Try with Hints

First hint

After the adding of the odd numbers, the total of the sequence increases by $836 – 715 = 121 = 11^2$.

Second Hint

Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\frac{715}{11} = 65$.

Final Step

Since the first, last, and middle terms are centered around the mean, then $65 \times 3 = 195$

Hence option B correct.

Categories

## Digits of number | PRMO 2018 | Question 3

Try this beautiful problem from the PRMO, 2018 based on Digits of number.

## Digits of number – PRMO 2018

Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.

• is 107
• is 70
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Numbers

Multiples

PRMO, 2018, Question 3

Higher Algebra by Hall and Knight

## Try with Hints

First hint

abccba (b is odd)

=a($10^5$+1)+b($10^4$+10)+c($10^3$+$10^2$)

=a(1001-1)100+a+10b(1001)+(100)(11)c

=(7.11.13.100)a-99a+10b(7.11.13)+(98+2)(11)c

=7p+(c-a) where p is an integer

Second Hint

Now if c-a is a multiple of 7

c-a=7,0,-7

hence number of ordered pairs of (a,c) is 14

Final Step

since b is odd

number of such number=$14 \times 5$=70.

Categories

## Smallest value | PRMO 2018 | Question 15

Try this beautiful problem from the PRMO, 2018 based on Smallest value.

## Smallest Value – PRMO 2018

Let a and b natural numbers such that 2a-b, a-2b and a+b are all distinct squares. What is the smallest possible value of b?

• is 107
• is 21
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Numbers

Multiples

PRMO, 2018, Question 15

Higher Algebra by Hall and Knight

## Try with Hints

First hint

2a-b=$k_1^2$ is equation 1

a-2b=$k_2^2$ is equation 2

a+b=$k_3^2$ is equation 3

Second Hint

adding 2 and 3 we get

2a-b=$k_2^2+k_3^2$

or, $k_2^2+k_3^2$=$k_1^2$ $(k_2<k_3)$

Final Step

For least ‘b’ difference of $k_3^2$ and $k_2^2$ is also least and must be multiple of 3

or, $k_2^2$=a-2b=$9^2$ and $k_3^2$=a+b=$12^2$

or, $k_3^2-k_2^2$=3b=144-81=63

or, b=21

or, least b is 21.

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## Length and Triangle | AIME I, 1987 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Length and Triangle.

## Length and Triangle – AIME I, 1987

Triangle ABC has right angle at B, and contains a point P for which PA=10, PB=6, and $\angle$APB=$\angle$BPC=$\angle$CPA. Find PC.

• is 107
• is 33
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Algebra

Triangles

AIME I, 1987, Question 9

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

First hint

Let PC be x, $\angle$APB=$\angle$BPC=$\angle$CPA=120 (in degrees)

Second Hint

Applying cosine law $\Delta$APB, $\Delta$BPC, $\Delta$CPA with cos120=$\frac{-1}{2}$ gives

$AB^{2}$=36+100+60=196, $BC^{2}$=36+$x^{2}$+6x, $CA^{2}$=100+$x^{2}$+10x

Final Step

By Pathagorus Theorem, $AB^{2}+BC^{2}=CA^{2}$

or, $x^{2}$+10x+100=$x^{2}$+6x+36+196

or, 4x=132

or, x=33.

Categories

## Algebra and Positive Integer | AIME I, 1987 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

## Algebra and Positive Integer – AIME I, 1987

What is the largest positive integer n for which there is a unique integer k such that $\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}$?

• is 107
• is 112
• is 840
• cannot be determined from the given information

### Key Concepts

Digits

Algebra

Numbers

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

## Try with Hints

First hint

Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

Second Hint

for 91n-104k<n+k, K>$\frac{6n}{7}$

and 0<91n-104k gives k<$\frac{7n}{8}$

Final Step

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.