Categories

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

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

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 .

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

Integers

Number Theory

Algebra

## Check the Answer

Answer: is 40.

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.

Categories

## 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

Series

Number Theory

Algebra

## Check the Answer

Answer: is 195.

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

## Try with Hints

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

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 $\frac{715}{11} = 65$.

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

Hence option B correct.

Categories

## 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

Angles

Algebra

Triangles

## Check the Answer

Answer: is 33.

AIME I, 1987, Question 9

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

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

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

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

Digits

Algebra

Numbers

## Check the Answer

Answer: is 112.

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

## Try with Hints

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

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

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

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

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

thus largest value of n=112.

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 $3 ?$

• $9$
• $7$
• $28$

Algebra

Equation

multiplication

## Check the Answer

Answer:$28$

PRMO-2017, Problem 1

Pre College Mathematics

## Try with Hints

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 ……….

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……..

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

## Distance and Spheres | AIME I, 1987 | Question 2

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

## Distance and Sphere – AIME I, 1987

What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)?

• is 107
• is 137
• is 840
• cannot be determined from the given information

Angles

Algebra

Spheres

## Check the Answer

Answer: is 137.

AIME I, 1987, Question 2

Geometry Vol I to Vol IV by Hall and Stevens

## Try with Hints

The distance between the center of the spheres is $\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}$

=$\sqrt{14^{2}+18^{2}+21^{2}}$=31

The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.

Categories

## Arithmetic Mean | AIME I, 2015 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean.

## Arithmetic Mean of Number Theory – AIME 2015

Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

• is 107
• is 431
• is 840
• cannot be determined from the given information

Inequalities

Algebra

Number Theory

## Check the Answer

Answer: is 431.

AIME, 2015, Question 12

Elementary Number Theory by David Burton

## Try with Hints

Each 1000-element subset ${ a_1, a_2,a_3,…,a_{1000}}$ of ${1,2,3,…,2015}$ with $a_1<a_2<a_3<…<a_{1000}$ contributes $a_1$ to sum of least element of each subset and set ${a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$. $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1$ ($k$ can be anything from $1$ to $a_1$ inclusive

Thus, the number of ways to choose the set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is equal to the sum. But choosing a set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is same as choosing a 1001-element subset from ${1,2,3,…,2016}$!

average =$\frac{2016}{1001}$=$\frac{288}{143}$. Then $p+q=288+143={431}$

Categories

## Algebra and Combination | AIME I, 2000 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebra and Combination.

## Algebra and combination – AIME 2000

In expansion $(ax+b)^{2000}$ where a and b are relatively prime positive integers the coefficient of $x^{2}$ and $x^{3}$ are equal, find a+b

• is 107
• is 667
• is 840
• cannot be determined from the given information

Algebra

Equations

Combination

## Check the Answer

Answer: is 667.

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

here coefficient of $x^{2}$= coefficient of $x^{3}$ in the same expression

then ${2000 \choose 1998}a^{2}b^{1998}$=${2000 \choose 1997}a^{3}b^{1997}$

then $b=\frac{1998}{3}$a=666a where a and b are relatively prime that is a=1,b=666 then a+b=666+1=667.

.

Categories

## Algebraic Equation | AIME I, 2000 Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

## Algebraic Equation – AIME 2000

Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, $x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$ then $z+\frac{1}{y}$=$\frac{m}{n}$ where m and n are relatively prime, find m+n

• is 107
• is 5
• is 840
• cannot be determined from the given information

Algebra

Equations

Integers

## Check the Answer

Answer: is 5.

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

## Try with Hints

here $x+\frac{1}{z}=5$ then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and $y=(29-\frac{1}{x}$) together gives 5-x=x$(29-\frac{1}{x}$) then x=$\frac{1}{5}$

then y=29-5=24 and z=$\frac{1}{5-x}$=$\frac{5}{24}$

$z+\frac{1}{y}$=$\frac{1}{4}$ then 1+4=5.

.