Categories
Math Olympiad

What is Parity in Mathematics ? 🧐

Parity in Mathematics is a term which we use to express if a given integer is even or odd. It is basically depend on the remainder when we divide a number by 2.

Parity can be divided into two categories – 1. Even Parity

2. Odd Parity

Even Parity : If we divide any number by 2 and the remainder is ‘0’,the parity is even or ‘0’ parity

Odd Parity : If we divide any number by 2 and the remainder is ‘1’,the parity is odd or ‘1’ parity. Till this we all know but let’s try to explore this by some of the problems.

Let’s understand it with the help of a few problems:

Problem 1: In how many ways can 10001 be written as the sum of two primes? (AMC 8,2011 Prob.28)

This problem can easily be solved using the ‘Parity’ concept.

The above rules have wide range of utility in Mathematics.

According to the problem the 10001 need to be expressed by the sum of two Prime Numbers.

10001 is odd number and to get the odd sum we need to add one odd number and one even number.This is the basic criteria of Parity.

Again the numbers should be Prime.

The only Even Prime Number is 2 but we cannot consider the other number to be 10001 – 2 = 9999 which is not a Prime Number.

Hence, We can’t express 10001 as the sum of the two Prime Numbers.

Problem 2 :

Suppose you have written the numbers 1 2 3 4 5 6 7 8 9 10.
You have to plug in ‘+’ or ‘-’ in between these numbers. You have the complete freedom to plug in anywhere.
The question is can the be sum zero? Ever?

Step 1: Let’s start by taking some numbers :

1 2 3 4 5 6 7 8 9 10

Step 2: Plug in the signs ‘+’ or ‘-‘.

1 + 2 + 3 – 4 + 5 – 6 – 7 + 8 – 9 – 10 = -17 ——– (1)

(You can take any signs its just an example)

Step 3: We need to make all the -ve numbers in the LHS of \(eq^n\) (1) into +ve numbers. It’s an easy calculation that if we add double the number (the numbers in negative) we will get the same number in positive, eg. -x + 2x = +x .

1 + 2 + 3 – 4 +8+ 5 – 6+12 – 7 +14+ 8 – 9+18 – 10 +20= -17+8+12+14+18+20

1+2+…………+10 = 55

Basically, (1 + 2 + 3 – 4 + 5 – 6 – 7 + 8 – 9 – 10) + (8+12+14+18+20) = 55

Initial Sum + Bunch of Positive numbers = Odd Number

Initial sum was = -17 again an odd number

Odd Number + Even Numbers(as double of any number is even) = Odd

But if the initial sum is ‘0’ which is an even number then it’s not possible.

As only , odd + even = odd .

Hence, It is not possible to be the initial sum as ‘0’.

Watch the Video – Parity in Mathematics:

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Categories
Singapore Math Olympiad

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

Check the Answer


But try the problem first…

Answer : 81

Source
Suggested Reading

Singapore Mathematical Olympiad

Challenges an Thrills – Pre – College Mathematics

Try with Hints


First hint

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…………..

Second Hint

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]

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Categories
Singapore Math Olympiad

Logarithm Problem From SMO, 2011 | Problem 7

Try this beautiful Logarithm Problem From Singapore Mathematics Olympiad, SMO, 2011 (Problem 7).

Logarithm Problem From SMO


  1. Let \(x=\frac {1}{\log_{\frac {1}{3}} \frac {1}{2}}\)+\(\frac {1}{\log_{\frac {1}{5}} \frac {1}{4}}\)+\(\frac {1}{\log _{\frac {1}{7}} \frac{1}{8}}\) Which of the following statements
    is true?
  • 1.5<x<2
  • 2<x<2.5
  • 2.5<x<3
  • 3<x<3.5
  • 3.5<x<4

Key Concepts


log function

Logarithmic

Inverse Exponentiation

Check the Answer


But try the problem first…

Answer: 3.5<x<4

Source
Suggested Reading

Singapore Mathematical Olympiad

Challenges and thrills – Pre – College Mathematics

Try with Hints


First hint

If you got stuck in this problem we can start from here:

\(x=\frac {1}{\log_{\frac {1}{3}} \frac {1}{2}}\)+\(\frac {1}{\log_{\frac {1}{5}} \frac {1}{4}}\)+\(\frac {1}{\log _{\frac {1}{7}} \frac{1}{8}}\)

If we refer too the basic properties of log we can find ,

x=\(\frac{\log \left(\frac{1}{3}\right)}{\log \left(\frac{1}{2}\right)}\)+\(\frac{\log \left(\frac{1}{5}\right)}{\log \left(\frac{1}{4}\right)}\)+\(\frac{\log \left(\frac{1}{7}\right)}{\log \left(\frac{1}{8}\right)}\)=\(\frac{-\log 3}{-\log 2}+\frac{-\log 5}{-\log 4}+\frac{-\log 7}{-\log 8}\)

Try the rest ………………………………..

Second Hint

\(\frac{-\log 3}{-\log 2}+\frac{-\log 5}{-\log 4}+\frac{-\log 7}{-\log 8}\)

so we can find

\(\frac {\log 3+ \log 5^{\frac {1}{2}}}+ \log 7^{\frac {1}{3}}{log 2}\)

= \(\frac {\log \sqrt {45} + log 7^{\frac {1}{3}}}{log 2}\) < \(\frac {\log \sqrt {65} + log 8^{\frac {1}{3}}}{log 2}\)

= \(\frac{3 \log 2+\log 2}{\log 2}=4\)

Try the rest …………………..

Final Step

Now let’s say ,

2x = \(2 \frac {log 3 + log 5^{\frac {1}{2}}+ log 7^{\frac {1}{3}}}{log 2}\)

=

\(\begin{array}{l}
\frac{\log (9 \times 5)+\log \left(49^{\frac{1}{3}}\right)}{\log 2}>\frac{\log \left(45 \times 27^{\frac{1}{3}}\right)}{\log 2} = \
\frac{\log (45 \times 3)}{\log 2}>\frac{\log (128)}{\log 2}=7
\end{array}\)

so x is greater than 3.5.

3.5 <x<4 is the correct answer.

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AMC 10 USA Math Olympiad

Combination of Equations | SMO, 2010 | Problem No. 7

Try this beautiful problem from Singapore Mathematical Olympiad, SMO, 2010 – Problem 7 based on the combination of equations.

Problem – Combination of Equations (SMO Entrance)


Find the sum of all the positive integers p such that the expression (x-p) (x – 13) + 4 can be expressed in the form (x+q) (x+r) for distinct integers q and r.

  • 26
  • 27
  • 16
  • 20

Key Concepts


Basic Algebra

Combination of Terms

Generator of a group

Check the Answer


But try the problem first…

Answer: 26

Source
Suggested Reading

Singapore Mathematical Olympiad

Challenges and Thrills – Pre College Mathematics

Try with Hints


First hint

If you got stuck in this problem start this problem using this hint :

Start with the given hint

(x-p) (x-13) +4 = (x+q)(x+r)

Let’s try to minimize the expression by taking x= -q

so , (-q -p)(-q -13) = -4 , which becomes (q+p) (q+13) = -4

As it is already given p and q are integers we can come up with many cases .

Try to find out the different cases we can have ………………………..

Second Hint

Starting after the last hint :

p+q = 4 and q+13 = -1 ………………………………..(1)

q+p = -4 and q +13 = 1……………..(2)

p+q = 2 and q+13 = -2 and …………………………(3)

p+q = -2 and q+13 = 2 …………………………(4)

For 1st case its simple calculation that we get q = -14 and p = 8

The initial expression becomes (x-p) (x-13) +4 = (x-14) (x-17)

For 2 nd case :

q= -12 and p = 8

so the initial expression becomes : (x-p)(x-13)+4 = (x-9)(x-12)

try the rest of the cases…………

Final Step

Now let’s talk about 3rd case ,

q= -15 , p = 17 and

hence (x – p) (x – 13) +4 = \(( x – 15)^2\) which is not true to this problem .

For last case , we obtain q = -11, q = 9 so the initial

(x- p) (x-13) +4 = \((x-11)^2\)

So p is 8 and 18 which add upto 8 +18 = 26 -(Answer)

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AMC 10 USA Math Olympiad

Trigonometry Simplification | SMO, 2009 | Problem 26

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Trigonometry Simplification.

Problem – Trigonometry Simplification (SMO Entrance)


If \(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ} = tan x^\circ \)

Find \( x^\circ \) ?

  • 12
  • 95
  • 46
  • 28

Key Concepts


Trigonometry

Geometry

Check the Answer


But try the problem first…

Answer: 95

Source
Suggested Reading

Singapore Mathematical Olympiad

Challenges and Thrill – Pre College Mathematics

Try with Hints


First hint

If you really got stuck into this sum we can start from here

\(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ}\)

= \(\frac {cos 100^\circ}{1-2sin 50^\circ cos 50^\circ}\)

Now let’s check with some basic values in trigonometry

\( Cos 2 A = cos^2 A – sin^2 A \) and

\(2 sin A cos A = sin 2 A\)

Now try the rest of the sum by using these two above mentioned values………………

Second Hint

Let’s continue from the last hint :

\( cos 100^\circ = cos^2 50^\circ – sin^2 50^\circ \)

\( 2 sin 25^\circ cos 25^\circ = sin 50^\circ\)

\(\frac {cos^2 50^\circ – sin^2 50^\circ}{2sin 50^\circ cos 50^\circ}\)

\(\frac {cos^2 50^\circ – sin^2 50^\circ }{(cos 50^\circ – sin 50^\circ)^2}\)

Using \(a^2 – b^2 = (a+b) (a-b)\) formula

\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ – sin 50^\circ}\)

Do the rest of the steps ……………..

Final Step

Starting from right after the last hint:

\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ – sin 50^\circ}\)

= \(\frac {1+ tan 50^\circ}{1-tan 50^\circ}\)

= \(\frac {tan 45^\circ + tan 50^\circ}{1-tan 45^\circ tan 50 ^\circ}\)

= \( tan 95^\circ\) – Answer

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Singapore Math Olympiad

Problem on Probability from SMO, 2012 | Problem 33

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2012 based on Probability.

Problem – Probability (SMO Entrance)


Two players A and B play rock – paper – scissors continuously until player A wins 2 consecutive games. Suppose each player is equally likely to use each hand – sign in every game . What is the expected number of games they will play?

  • 12
  • 15
  • 16
  • 20

Key Concepts


Probability

Permutation Combination

Check the Answer


But try the problem first…

Answer: 12

Source
Suggested Reading

Singapore Mathematical Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


First hint

We can start with a general case :

Two players are playing a series of games of Rock – Paper – scissors. There are a total of K games played. Player 1 has a sequence of moves denoted by string A and similarly player 2 has string B. If any player reaches the end of their string, they move back to the start of the string. The task is to count the number of games won by each of the player when exactly K games are being played.

To start with this particular problem let’s set an expectation as k.

If A doesnot win , so the probability will be = \(\frac {no of event}{total event} = \frac {2}{3}\)

so game will be restarted.

Try to find out the rest of the cases…………..

Second Hint

Continue from the 1st hint :

Again case 2: If A wins at first and then losses so again the probability will be \(\frac {1}{3} \times \frac {2}{3}\). again new game will start.

Last case : A wins in consecutive two games the probability will be \(\frac {1}{3} \times \frac{1}{3} \)

Do the rest of the sum……..

Final Step

The total number of games that they will play :

K = \(\frac {2}{3} \times (K+1) \times \frac {2}{9} (K+2) \times \frac {1}{9} \times 2 \)

Now solve this equation and at the end E will be 12. [Check it yourself]

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Singapore Math Olympiad

Problem on Permutation | SMO, 2011 | Problem No. 24

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2011 based on Permutation.

Permutation Problem (SMO Entrance)


A \(4 \times 4\) Sudoku grid is filled with digits so that each column , each row and each of the four \( 2 \times 2\) sub grids that composes the grid contains all of the digits from 1 to 4. For example

Sudoku - Permutation Problem
  • 288
  • 155
  • 160
  • 201

Key Concepts


Permutations & Combinations

Sudoku

Set Theory

Check the Answer


But try the problem first…

Answer: 288

Source
Suggested Reading

Singapore Mathematical Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


First hint

If you really get stuck in this problem here is the first hint to do that:

At 1st let’s consider the sub grids of \( 2 \times 2\) filled with 1-4 ( 1, 2 , 3 ,4)

If a,b,c,d are all distinct , and there are no other numbers to place in x . If {a,b} = {c,d} then again a’,b’,c,d are all distinct , and no other number can be possible for x’.

We need to understand that the choices we have ,

{a,a’} = {1,2} , {b,b’} = {3,4}, {c,c’} = {2,4} and {d,d’} = {1,3}

Among these choices \( 2^4 = 16 \) choices 4 of them are impossible – {a,b} = {c,d} = {1,4} or {2,3} and

{a,b} = {1,4} and {c,d} = {2,3} and {a,b} = {2,3} and {c,d} = {1,4}

Try rest….

Second Hint

Now for each remaining case a’,b’,c’ and d’ are uniquely determined so

{x} = {1,2,3,4} – {a,b} \(\cup\) {c,d}

{y} = {1,2,3,4} – {a,b} \(\cup\) {c’,d’}

{x’} = {1,2,3,4} – {a’,b’} \(\cup\) {c,d}

{y’} = {1,2,3,4} – {a’,b’} \(\cup\) {c’,d’}

Final Step

In final hint :

There are 4! = 24 permutation in the left top grid we can find. So total 12 * 24 = 288 possible 4\(\times\) 4 Sudoku grids can be found.

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Categories
Singapore Math Olympiad

Application of Pythagoras Theorem | SMO, 2010 | Problem 22

Try this problem from the Singapore Mathematics Olympiad, SMO, 2010 based on the application of the Pythagoras Theorem.

Application of Pythagoras Theorem- (SMO Test)


The figure below shows a circle with diameter AB. C ad D are points on the circle on the same side of AB such that BD bisects \(\angle {CBA}\). The chords AC and BD intersect at E. It is given that AE = 169 cm and EC = 119 cm. If ED = x cm, find the value of x.

  • 65
  • 55
  • 56
  • 60

Key Concepts


Circle

Pythagoras Theorem

2D – Geometry

Check the Answer


But try the problem first…

Answer: 65

Source
Suggested Reading

Singapore Mathematical Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


First hint

If you get stuck in this problem this is the first hint we can start with :

As BE intersect \(\angle {CBA}\) we have \(\frac {BC}{BA} = \frac {EC}{EA} = \frac {119}{169}\)

Thus we can let BC = 119 y and BA = 169 y .

Since \(\angle {BCA} = 90 ^\circ\).

Then try to do the rest of the problem ………………………………………………

Second Hint

If we want to continue from the last hint we have :

Apply Pythagoras Theorem ,

\(AB ^2 = AC^2 + BC ^2\)

\((169y)^2 = (169 + 119)^2 + (119y)^2\)

\(y^2 (169-119)(169+119) = (169+119)^2\)

\(y^2 = \frac {169+119}{169-119} = \frac {144}{25}\)

\(y = \frac {12}{5}\)

Final Step

In the last hint:

Hence , from triangle BCE , we have BE = \(\sqrt{119^2 + (119y)^2} = 119 \times \frac {13}{5}\)

Finally , note that \(\triangle {ADE}\) and \(\triangle {BCE}\) are similar , so we have

ED = \(\frac {AE \times CE}{BE} = \frac {169 \times 119}{119 \times \frac {13}{5}} = 65 \) cm .

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Categories
AMC 10 USA Math Olympiad

Problem on Trigonometry | SMO, 2008 | Problem – 22

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.

Problem on Trigonometry | SMO, 2008 |Problem 22

Find the value of \(\frac {tan 40^\circ tan 60^\circ tan 80^\circ}{tan40^\circ + tan 60^\circ + tan 80^\circ}\)

  • 1
  • 15
  • 6
  • 0

Key Concepts


Trigonometry

Tan Rule

Check the Answer


But try the problem first…

Answer: 1

Source
Suggested Reading

Singapore Mathematical Olympiad, 2008

Challenges and Thrills – Pre College Mathematics

Try with Hints


First hint

If you got stuck in this sum how to get started you can start by consider a general case where \(40^\circ = A\) , \(60^\circ = B\) and \(80^\circ = C\).

So , A+B+C = \( 180 ^\circ\)

\( A+B = 180^\circ – C\)

(tan (A+B) = tan \(180^\circ – C)\)……………………….(1)

Now try to implement the basic formula and try to do this sum………………

Second Hint

In this we can continue from the last hint:

the formula of tan (A + B) = \(\frac {tan A + tan B}{1- tan A . tan B}\)

From the equation (1) …….

tan (A+B) = tan (180 – C)

\(\frac {tan A + tan B}{1- tan A . tan B} = tan (180^\circ – c)\)

(frac {tan A + tan B}{1- tan A . tan B} = -tan C )

Now just rearrange this expression and you will get the final answer……………..

Final Step

Here is the final solution:

{tan A + tan B} = -tan C {1- tan A . tan B}

tan A + tan B = – tan C + tan C tan A tan B

tan A + tan B + tan C = tan A tan B tan C

\(\frac {tan A tan B tan C}{tan A + tan B + tan C} = 1\)

Which is the given question. It can be a proof also……………….

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Categories
Singapore Math Olympiad

Functional Equations Problem | SMO, 2012 | Problem 33

Try this beautiful Problem from Singapore Mathematics Olympiad, 2012 based on Functional Equations.

Problem – Functional equations (SMO Test)


Let L denote the minimum value of the quotient of a 3- digit number formed by three distinct divided by the sum of its digits.Determine \(\lfloor 10L \rfloor \).

  • 105
  • 150
  • 102
  • 200

Key Concepts


Functional Equation

Max and Min Value

Check the Answer


But try the problem first…

Answer: 105

Source
Suggested Reading

Singapore Mathematical Olympiad, 2012

Challenges and Thrills – Pre – College Mathematics

Try with Hints


First hint

If you got stuck at first only here is the hint to begin with :

Anyway a three digit number we can be expressed as 100x + 10 y +z depending on the place values. and if we do minimize it :

F(x y z) = \(\frac {100x + 10y + z}{x + y + z}\)

Lets consider that for distinct x , y , z, F(x , y , Z) has the minimum value when x<y<z.

Again we can assume,

\( 0 < a < b < c \leq 9\)

Note ,

F(x,y,z) = \(\frac {100 x + 10 y + z }{x +y + z}\) = 1 + \(\frac {99 x + 9 y }{x+y+z}\)

Try the rest of the sum……………

Second Hint

From the last hint we can say

F(x y z ) is minimum when c = 9 (say)

F(x y 9) = 1+ \(\frac {99x +9y }{x+y+9} = 1 + \frac {9(x + y + 9) + 90 x – 81}{x+y +9} = 10 + \frac {9(10x -9)}{x+y +9}\)

Try to do for the next case for minimum value when b = 8………………

Final Step

In the last hint we can do the next step which is b= 8:

F(x 8 9) = 10 + \(\frac {9(10x -9)}{x+17}= 10 + \frac {90(a+17)- 1611}{x + 17} = 100 – \frac {1611}{x +17} \)

which has the minimum value of x = 1 and so 10 L = 105.(answer)

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