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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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AMC-8 Math Olympiad USA Math Olympiad

Time & Work Problem | PRMO-2017 | Problem 3

Try this beautiful problem from PRMO, 2017 from Arithmetic based on Time & Work.

Time & Work Problem | PRMO | Problem 3


A contractor has two teams of workers: team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins A after four days. Team A withdraws after two more days. For how many more days should team B work to complete the job?

  • \(24\)
  • \(16\)
  • \(22\)
  • \(18\)

Key Concepts


Arithmetic

Unitary process

Work done

Check the Answer


Answer:\(16\)

PRMO-2017, Problem 3

Pre College Mathematics

Try with Hints


At first we have to find out A’s 1 days work and B’s 1 days work.next find out A and B both together 1 day’s work .

Can you now finish the problem ……….

Team A completes job in 12 days and Team B completes job in 36 days

1 day work of team A =\(\frac{1}{12}\)

1 day work of team B=\(\frac{1}{36}\)

1 day work of team A and team B (when they both work together \(\frac{1}{12} +\frac{1}{36}\)=\(\frac{1}{9}\)

Now according to question,
Let more number of days should team B works to complete the job be x days

\(4 \times \frac{1}{12} +2 \times \frac{1}{9} + x \times \frac{1}{36}=1\)

\(\Rightarrow x=16\)

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AMC 10 USA Math Olympiad

Arithmetic Sequence | AMC 10B, 2003 | Problem No. 24

Problem – Arithmetic Sequence (AMC 10)


The first four terms in an arithmetic sequence are  \( x+y\) , \( x-y\) xy , \(\frac {x}{y}\) in that order. What is the fifth term?

  • \(\frac{12}{110}\)
  • \(\frac {123}{40}\)
  • \(\frac {16}{17}\)
  • 20

Key Concepts


Arithmetic Sequence

Series and Sequence

Algebra

Check the Answer


Answer: \(\frac {123}{40}\)

American Mathematics Competition

Challenges and Thrills – Pre – College Mathematics

Try with Hints


Here is the first hint to start this sum:

There is a very easy method to do this sum

At first we can try to find the difference between two consecutive terms which is

\( (x-y) – (x+y) = -2y \)

So after that we can understand the third and forth terms in terms of x and y.

They can be : \( ( x-3y ) \) and \( ( x – 5y )\)

Now try to do rest of the sum………………………………..

If you got stuck after the first hint you can use this :

Though we from our solution we find the other two terms to be \((x-3y)\) and \((x-5y)\)

but from the question we find that the other two terms are \(xy\) and \(\frac {x}{y}\)

So both are equal.Thus ,

\(xy = x – 3y\)

\( xy – x = – 3y \)

\( x (y – 1) = -3y \)

\( x = \frac {-3y}{ y – 1} \) ……………………………(1)

Again , similarly

\(\frac {x}{y} = x -5y \)

Now considering the equation (1) we can take the value of \(\frac {x}{y}\)

\(\frac {-3}{y – 1}= \frac {-3y}{y -1} – 5y \) …………………….(2)

\( -3 = -3y – 5y(y-1) \)

\( 0 = 5y^2 – 2y – 3\)

\( 0 = ( 5y +3)(y-1)\)

\( y = – \frac {3}{5} , 1\)

We are almost there with the answer. Try to find the answer…..

Now from the last hint we find the value of \( y = – \frac {3}{5} , 1\)

But we cannot consider the value of y to be 1 as the 1st and 2 nd terms would be \(x+1\) and \(x-1)\) but last two terms will be equal to x .

So the value of y be \(- \frac {3}{5}\) and substituting the value of y in either \(eq^n\) (1) or \(eq^n\) (2) we get x = -\(\frac {9}{8}\)

so , \(\frac {x}{y} -2y = \frac {9.5}{8.3} + \frac {6}{5} \)

= \(\frac {123}{40} \) (Answer )

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AMC 10 Australian Math Competition USA Math Olympiad

Statistics Problem – Australian Mathematics Competition, 2014

Problem No – 21 – Australian Mathematics Competition – 2014


Here is a problem based on Statistics from Australian Mathematics Competition, 2014.

In a competition between four people, Sally scored twice as many points as Brian and 30 points more than Corrie. Donna scored 50 points more than Brian. Which of the following statements is definitely true?

  • Sally won the competition.
  • Brian came last in the competition.
  • Donna won the competition.
  • Corrie beat Brian.
  • Sally and Donna together scored more than Brian and Corrie.

Key Concepts


Mathematical Analysis

Statistics

Arithmetic

Check the Answer


Answer: Sally and Donna together scored more than Brian and Corrie.

Australian Mathematics Competition – Upper Primary Division – 2014 – UP 21

Statistics 10th Edition – Robert S. Witte and John S. Witte

Try with Hints


For first hint we can use a table form with some possibilities :

statistics problem - solution

So in the previous hint from the table its clear that the 1st , 2nd and 4th options are not correct. Now if we go with the higher numbers like 90,100 etc then there will be a change with the values.Try to find it out using a table.

I guess you have noticed the differences in values let try to do that……

statistics problem - solution

So can understand from this table that the 3rd option is also not correct .

We are told that Sally scored 30 points more than Corrie and Donna scored 50 points more than Brian, and so together Sally and Donna always scored 80 points more than Corrie and Brian

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AMC 10 USA Math Olympiad

Missing Numbers – Australian Mathematics Competition

Problem – Hidden Faces of Dice (AMC Middle Primary)


Let’s solve a problem based on missing numbers from the hidden faces of dice.

Three standard dice are sitting next to each other as shown in the diagram. There are 7 faces visible. How many dots are hidden on the other 11 sides?

Missing numbers - dice
  • 26
  • 36
  • 41
  • 63

Key Concepts


Dice Problem

Missing Values

Arithmetic

Check the Answer


Answer: 41

Australian Mathematics Competition – Middle Primary Division

Basic Arithmetic by Robert Moon

Try with Hints


This one is very easy but if you really need any hints at first take a dice and try to understand the number of dots in each sides .

So for first dice : we can see 4,5 and 1. So the no of dots which are not seen 6,2 and 3

For Second Dice : The visible dots are 2 and 6 . So the dots in sides which are not visible are 4,3,1 and 5.

For Third Dice : The Visible dots are 3 and 1 .So the dots which are not visible 2,4,5 and 6.

I think you have already got the answer but if you really need this last step :

Now add all the number of dots from the invisible sides:

2+4+5+6+4+3+1+5+6+2+3 = 41 .

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Math Olympiad Themes in Mathematics

Shapes and Measure 1 – let us divide!

This is a part of Thousand Flowers Pre-Olympiad lectures at Cheenta. It is an experiment to teach percentage arithmetic, euclidean geometry, rational and irrationals and computational software tools in an interdisciplinary manner. We extensively used GeoGebra and plan to use Python later in this sequence of discussion. The exposition is conversational (in the spirit of Tarasov’s Pre-Calculus). As a method of pedagogy, dialectics is subtly employed to enhance student’s grasp of the subject. 

Teacher: Cent is a French word. It means 100. “Percentage” essentially means: divide the whole quantity into 100 parts.

Student: But why should we do it in the first place?

T: There are two main goals:

  • compare portions with the whole.
  • measure large (or small) pieces in terms of the whole.

The key idea is, we have to fix the whole first.

S: What is the whole?

T: It depends on the context. For example, the whole could be the monthly salary of a person. Suppose monthly salary of Mr. Bakshi is Rs. 20,000. Then that is the whole quantity. If he wants to buy a mobile phone worth Rs. 12,000, he may want to know, what portion of the salary is being spent on this single purchase.

So he divides whole quantity (Rs. 20,000) into 100 parts. Each part is Rs. 200.

Rs. 12,000 is made up of 60 such parts. ( (60 \times 200 = 12,000 ) ). Clearly, he is spending 60 out of 100 parts on this purchase and we say it is 60% of the income.

S: I see. So we wish to compare the partial expense with the total money available.

T: Exactly. This may change according to the context. Here is a problem, that you may try for fun:

Problem: Consider the cricket career of Sachin Tendulkar. Find from the internet, the number one-day matches that he played. Next, find the number of matches he has scored 50 or more. If whole quantity = the total number of matches = 100 parts, then in how many parts he scored fifty or more.

S: What is the other use? How can we measure by dividing the whole into 100 parts?

T: In order to do this, you have to first fix a reference length. It could be any length (there is no need to use a ‘ruler’; in fact, we will be building a ruler). Take any length and declare that length as the whole quantity = 100 units.

Suppose AB is that length.

Now if you wish to measure the length XY (suppose larger than AB), one way to do it is the following:

  • take a compass,
  • put the needle at A and pencil at B. This measures the length AB. (This is why compasses are used)
  • without changing the aperture, put the needle at X and make a mark on the line XY
  • repeat this as many times as you can

As an example, consider the following picture.

compass and measure

Remember that we declared the length of AB to be 100. Can you estimate the length of XY from this picture?

S: I cannot say exactly. It seems to me, that XY is more than 300 but definitely less than 400.

T: Excellent. We could ‘cut’ three times with the length of AB. And there is some more. How do you think we can measure this remaining portion after the third cut?

S: Taking a cue from what you said earlier, we can divide the whole (=AB) into 100 parts, and try to fit in as many parts as possible in the remaining portion.

T: Precisely! Suppose we can fit in 72 of these 100 parts in the remaining portion after the third cut. But even after that, some tiny portion remains excess. What should we do?

S: Ah!  I am out of my wits. What can be done here?

T: Well, we may further divide the hundredth part of AB into 100 parts. Now, each tiny part is 1/10000th part of AB. We will try to fit in as many of these super tiny lengths into the excess portion as possible.

S: I see. Since these pieces are super thin, we will be able to fit in several pieces of this in the excess portion.

T: Right.

S: But will that cover that excess portion entirely?

T: This is an excellent question. If it does not, we will have to divide 1/10000 th part of AB into 100 parts again and try one more. But there is no guarantee, that if we keep of doing this, we will be able to exactly fill in the entire length of XY (with no super little part remaining uncovered).

In fact, one of the key discoveries of the ancient Greeks was to find such a length, which can never be exactly covered by these pieces (no matter how many times we split them into hundred parts and try to fill up the excess portion).

S: Wow! I wonder, what length will that be.

T: Actually there are many such lengths. Create a right triangle, with the length of legs = AB. Then the hypotenuse is one such length. Of course, we will need to prove this claim. We will come back to this idea. But even before doing all these, we have to resolve one issue: how to split AB into a hundred equal parts.

In fact, let’s handle a much simpler problem: how to split AB into two equal parts (using a compass and an unmarked straight edge).

 

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Math Olympiad

Prime factor of last page – Pre RMO 2018 Problem 1 Discussion

A book is published in three volumes, the pages being numbered from 1 onwards. The page numbers are continued from the first volume to the third. The number of pages in the second volume is 50 more than in the first volume, and the numberpages in the third volume is one and a half times that in the second. The sum of the page numbers on the first pages of the three volumes is 1709. If n is the last page number. What is the largest prime factor of n?
Suppose the number of pages is the First Volume is t. It’s first page is numbered 1. Then the second volume has t+50 pages. Its first page is numbered t+1. Finally, the third volume has \( \frac{3}{2} \times ( t + 50) \) pages. Its first page is (t + t + 50 + 1 =) 2t + 51.
Sum of the page numbers of first pages of three volumes is: 1 + t +1 + 2t + 51 = 3t + 51. 3t + 51 = 1709 This implies t= 552
The total number of pages in three volumes is \( t + t + 50 + \frac{3}{2} \times (t+50) \\ = \frac {4t + 100 + 3t + 150}{2} \\ = \frac { 7 \times 552 + 250}{2} \\ = 2057 = 11^2 \times 17 \). Hence the greatest prime factor is 17.

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