# Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

## Sequence and fraction - AIME I, 2000

A sequence of numbers $$x_1,x_2,....,x_{100}$$ has the property that, for every integer k between 1 and 100, inclusive, the number $$x_k$$ is k less than the sum of the other 99 numbers, given that $$x_{50}=\frac{m}{n}$$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 173
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

Let S be the sum of the sequence $$x_k$$

given that $$x_k=S-x_k-k$$ for any k

$$100S-2(x_1+x_2+....+x_{100})=1+2+....+100$$

Second Hint

$$\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050$$

$$\Rightarrow S=\frac{2525}{49}$$

for $$k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$$

Final Step

$$\Rightarrow x_{50}=\frac{75}{98}$$

$$\Rightarrow m+n$$=75+98

=173.

# Functional Equation Problem from SMO, 2018 - Question 35

Try to solve this problem number 35 from Singapore Mathematics Olympiad, SMO, 2018 based on Functional Equation.

## Problem - Functional Equation (SMO Entrance)

Consider integers $${1,2, \ldots, 10}$$. A particle is initially -at 1 . It moves to an adjacent integer in the next step. What is the expected number of steps it will take to reach 10 for the first time?

• 82
• 81
• 80
• 79

### Key Concepts

Functional Equation

Equation

Challenges an Thrills - Pre - College Mathematics

## Try with Hints

If you got stuck into this problem we can start taking an expected number of steps to be $$g_{n}$$. We need to remember at first the particle was in 1 then it will shift to the next step so for n no of position we can expressed it as n and n -1 where n = 2,3,4,........,100.

Now try the rest..............

Now let's continue after the last hint ............

Then $$g_{n+1} = \frac {1}{2} (1+g_{n} + g_{n+1} )+ \frac {1}{2}$$

which implies , $$g_{n+1} = g_{n} + 2$$

Now we know that,$$g_{2} = 1$$. Then $$g_{3} = 3$$, $$g_{4}= 5$$,..................,$$g_{10}=17$$

$$g = g_{2}+g_{3}+g_{4}+....................+g_{10} = 1+3+.....................+17 = 81$$[ Answer]

# Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

## Sequence and greatest integer - AIME I, 2000

Let S be the sum of all numbers of the form $$\frac{a}{b}$$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $$\frac{S}{10}$$.

• is 107
• is 248
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

We have 1000=(2)(2)(2)(5)(5)(5) and $$\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3$$

Second Hint

sum of all numbers of form $$\frac{a}{b}$$ such that a and b are relatively prime positive divisors of 1000

=$$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$$

Final Step

$$\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times$$ $$\frac{(5^{-3})(5^{7}-1)}{5-1}$$

=2480 + $$\frac{437}{1000}$$

$$\Rightarrow [\frac{s}{10}]$$=248.

# 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

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.

# 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

PRMO II, 2019, Question 27

Higher Algebra by Hall and Knight

## Try with Hints

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

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$$\sqrt{3}$$

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

# Length of side of Triangle | PRMO II 2019 | Question 28

Try this beautiful problem from the Pre-RMO II, 2019, Question 28, based on Length of side of triangle.

## Length of side of triangle - Problem 28

In a triangle ABC, it is known that $$\angle$$A=100$$^\circ$$ and AB=AC. The internal angle bisector BD has length 20 units. Find the length of BC to the nearest integer, given that sin 10$$^\circ$$=0.174.

• is 107
• is 27
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

PRMO II, 2019, Question 28

Higher Algebra by Hall and Knight

## Try with Hints

First hint

given, BD=20 units

$$\angle$$A=100$$^\circ$$

AB=AC

In $$\Delta$$ABD

$$\frac{BD}{sinA}=\frac{AD}{sin20^\circ}$$

or, $$\frac{BD}{sin100^\circ}=\frac{AD}{sin20^\circ}$$

or, 20=$$\frac{AD}{2sin10^\circ}$$ or, AD=40sin10$$^\circ$$=6.96

Second Hint

In $$\Delta$$BDC

$$\frac{BD}{sin40^\circ}=\frac{BC}{sin120^\circ}=\frac{CD}{sin20^\circ}$$

or, CD=$$\frac{20}{2cos20^\circ}$$=$$\frac{20}{2 \times 0.9394}$$=10.65

Final Step

since BD is angle bisector

$$\frac{BC}{AB}=\frac{CD}{AD}$$

or, BC=$$\frac{AB \times CD}{AD}$$=$$\frac{17.6 \times 10.65}{6.96}$$

=26.98=27.

# Acute angled Triangle | PRMO II 2019 | Question 29

Try this beautiful problem from the Pre-RMO II, 2019, Question 29, based on Acute angled triangle.

## Acute angled triangle - Problem 29

Let ABC be a acute angled triangle with AB=15 and BC=8. Let D be a point on AB such that BD=BC. Consider points E on AC such that $$\angle$$DEB=$$\angle$$BEC. If $$\alpha$$ denotes the product of all possible val;ues of AE, find[$$\alpha$$] the integer part of $$\alpha$$.

• is 107
• is 68
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

PRMO II, 2019, Question 29

Higher Algebra by Hall and Knight

## Try with Hints

First hint

The pairs $$E_1$$,$$E_2$$ satisfies condition or $$E_1$$=intersection of CBO with AC and $$E_2$$=intersection of $$\angle$$bisector of B and AC

since that $$\angle DE_2B$$=$$\angle CE_2B$$ and for $$E_1$$$$\angle BE_1C$$=$$\angle$$BDC=$$\angle$$BCD=$$\angle BE_1D$$

or, $$AE_1.AC$$=$$AD.AB$$=$$7 \times 15$$

$$\frac{AE_2}{AC}$$=$$\frac{XY}{XC}$$

(for y is midpoint of OC and X is foot of altitude from A to CD)

Second Hint

$$\frac{XD}{DY}=\frac{7}{8}$$ and DY=YC

or, $$\frac{XD+DY}{XC}$$=$$\frac{15}{7+8+8}$$=$$\frac{15}{23}$$

or, $$\frac{XY}{XC}=\frac{15}{23}$$

or, $$\frac{AE_2}{AC}$$=$$\frac{15}{23}$$

or, $$AE_1.AE_2$$=$$\frac{15}{23}(7.15)$$=$$\frac{225 \times 7}{23}$$

Final Step

$$[\frac{225 \times 7}{23}]$$=68.

# Area of a part of circle | PRMO 2017 | Question 26

Try this beautiful problem from the Pre-RMO, 2017, Question 26, based on Area of part of circle.

## Area of part of circle - Problem 26

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

• is 107
• is 75
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

PRMO, 2017, Question 26

Higher Algebra by Hall and Knight

## Try with Hints

First hint

A=2[$$\frac{1}{2} \times 25 \times \theta$$]+$$\frac{1}{2} \times 3 \times 8$$+$$\frac{1}{2} \times 4 \times 6$$

where $$\theta=[\pi-(\theta_1+\theta_2)]=[\pi-(tan^{-1}\frac{4}{3}+tan^{-1}\frac{3}{4})]$$

Second Hint

or, $$\theta=\frac{\pi}{2}$$

or, A=24+$$\frac{25\pi}{2}$$

or, A=$$\frac{48+25\pi}{2}$$

Final Step

(m+n+k)=(48+2+25)=75

# Number of points | TOMATO B.Stat Objective 713

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Number of points.

## Numbers of points (B.Stat Objective Question)

The number of points in the rectangle {(x,y)|$$-10 \leq x \leq 10 and -3 \leq y \leq 3$$} which lie on the curve $$y^{2}=x+sinx$$ and at which the tangent to the curve in parallel to x axis, is

• 0
• 4
• 53361
• 5082

### Key Concepts

Equation

Roots

Algebra

B.Stat Objective Problem 713

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

here we see two graphs $$y^{2}=x+sinx$$ and rectangle region

Second Hint

four such points here in which tangent is parallel to x axis

Final Step

or, number of required points=4.

# Derivative Problem | TOMATO BStat Objective 764

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Derivative.

## Derivative Problem (B.Stat Objective Question )

If y=$$3^\frac{sinax}{cosbx}$$, then $$\frac{dy}{dx}$$ is

• $$3^\frac{sinax}{cosbx}\frac{acosaxcosbx-bsinaxsinbx}{cos^{2}bx}log3$$
• $$3^\frac{sinax}{cosbx}\frac{acosaxcosbx+bsinaxsinbx}{cos^{2}bx}log3$$
• $$3^\frac{sinax}{cosbx}log3$$
• $$3^\frac{sinax}{cosbx}$$

### Key Concepts

Equation

Derivative

Algebra

Answer:$$3^\frac{sinax}{cosbx}\frac{acosaxcosbx+bsinaxsinbx}{cos^{2}bx}log3$$

B.Stat Objective Problem 764

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

$$y=3^\frac{sinax}{cosbx}$$

or, $$\frac{dy}{dx}$$=$$3^{\frac{sinax}{cosbx}}log_{e}{3}\frac{d}{dx}\frac{sinax}{cosbx}$$

Second Hint

or, $$\frac{dy}{dx}$$=$$3^{\frac{sinax}{cosbx}}log_{e}{3}{\frac{cosbxcosaxa-sinax(-sinbx)b}{cos^{2}bx}}$$

Final Step

or, $$3^\frac{sinax}{cosbx}\frac{acosaxcosbx+bsinaxsinbx}{cos^{2}bx}log3$$