Math Kangaroo (Benjamin) 2014 Problem No 24

Try this beautiful Problem based on simple Algebra appeared in Math Kangaroo (Benjamin) 2014 Problem 24.

Math Kangaroo (Benjamin) 2014 | Problem No 24

Grandma gives 180 marbles to her ten grandchildren. No two children get the same amount of marbles. Anna gets the most. What is the minimum number of marbles that Anna could get?

Key Concepts

Arithmetic

Equation solving

Algebra

Suggested Book | Source | Answer

Algebra by Gelfand

Math Kangaroo (Benjamin), 2014

$23$

Try with Hints

Here

Number of children is $10$.
Number of marbles is $180$.
And Anna gets the most and no 2 children gets the same number of marbles.

Let us assume that Anna could get $x$ marbles and also the other 9 children receiving 1 less each step. Apply the condition to construct the equation

The equation will be-

$x$+$x$-$1$+$x$-$2$+$x$-$3$+$x$-$4$+$x$-$5$+$x$-$6$+$x$-$7$+$x$-$8$+$x$-$9$=$180$

Now solve for $x$.

So, Anna can have $23$ marbles.

AMC 10A 2021 Problem 9 | Factorizing Problem

Try this beautiful Problem based on Factorizing Problem from AMC 2021 Problem 9.

Factorizing Problem: AMC 10A 2021 Problem 9

What is the least possible value of $(x y-1)^{2}+(x+y)^{2}$ for real numbers $x$ and $y$ ?

0
$\frac{1}{2}$
$\frac{1}{4}$
1
2

Key Concepts

Expansion of Polynomial

Factorization

Suggested Book | Source | Answer

AMC 10A 2021 Problem 9

Try with Hints

Expand the expression.

So, we get that the expression is $x^{2}+2 x y+y^{2}+x^{2} y^{2}-2 x y+1$ or $x^{2}+y^{2}+x^{2} y^{2}+1$.

Then the minimum value for this is 1 , which can be achieved at $x=y=0$. .

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

Check the Answer

Answer: is 173.

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

Try with Hints

Let S be the sum of the sequence $x_k$

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

taking k=1,2,....,100 and adding

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

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

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

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

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

$\Rightarrow m+n$=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?

Key Concepts

Functional Equation

Equation

Check the Answer

Answer : 81

Singapore Mathematical Olympiad

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

Check the Answer

Answer: is 248.

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

Try with Hints

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

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

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

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

=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

Check the Answer

Answer: is 31.

PRMO II, 2019, Question 26

Higher Algebra by Hall and Knight

Try with Hints

x= length of line segment

and by cosine law on triangle of side x, 1, 5 and 120 (in degrees) as one angle gives

$x^2=5^2+1^2-2 \times 5 \times 1 cos 120^\circ$

$=25+1+5$

or, x=$\sqrt{31}$

or, N=31

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

Check the Answer

Answer: is 36.

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

Check the Answer

Answer: is 27.

PRMO II, 2019, Question 28

Higher Algebra by Hall and Knight

Try with Hints

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

finding the length of the side of triangle

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

So, AD+CD=AC=AB=17.6

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

Check the Answer

Answer: is 68.

PRMO II, 2019, Question 29

Higher Algebra by Hall and Knight

Try with Hints

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)

$\frac{XD}{DY}=\frac{7}{8}$ 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, $AE_1.AE_2$=$\frac{15}{23}(7.15)$=$\frac{225 \times 7}{23}$

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

Check the Answer

Answer: is 75.

PRMO, 2017, Question 26

Higher Algebra by Hall and Knight

Try with Hints

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

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

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

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

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

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