Monday, December 16, 2013

Dimensional Change

There are lots of questions on dimensional change and this is a very common one.

Make sure you understand the relationship among linear, 2-D (area) and 3-D (volume) ratio.

There are many similar triangles featured in the image on the left.
Each of the two legs of the largest triangles is split into 4 equal side lengths.


Question : What is the area ratio of the sum of the two white trapezoids to the largest triangle? 
\(\dfrac {\left( 3+7\right) } {16}=\dfrac {10} {16}=\dfrac{5}{8}\)  

Question: If the area of the largest triangles are 400 square units, what is the area of the blue-colored trapezoid?
\(\dfrac {5} {16}\times 400\) =125 square units 

Again, each of the two legs are split into three equal segments. 

The volume ration of the cone on the top to the middle frustum to the 
bottom frustum is 1 : 7 : 19. 
Make sure you understand why.


Thursday, December 12, 2013

2014 Mathcounts State Prep : Folding Paper Questions

Folding paper questions are not too bad so here are two examples:

Question #1:  The two side lengths of rectangle ABCD is "a" and "b". If you fold along EF and the point B now converges on point D, what is the length of \(\overline {EF}\)?

Solution I :
Let the length of \(\overline {FC}\)  be x and the length of \(\overline {FD}\)  is thus a - x.

Using Pythagorean theorem you'll easily get x (the \( x^{2}\) part cancel each other out) and from there get the length of a - x.

\(\overline {HD}\) = \(\dfrac {1} {2}\) of the hypotenuse. (Use Pythagorean theorem or Pythagorean triples to get that length,)

Again, using Pythagorean theorem you'll get the length of \(\overline {HF}\). Times 2 to get \(\overline {EF}\).

Solution II : 
After you find the length of x, use \(\overline {EG}\), which is a - 2x and b as two legs of the right triangle EGF, you can easily get \(\overline {EF}\). (Pythagorean theorem)

Question #2: 

What about this time you fold B to touch the other side.
 What is the length of EF? 

This one is not too bad.

Do you see there are two similar triangles?

Just make sure you use the same corresponding sides to get the desired

Tuesday, December 10, 2013

Answer to one mathleague quesiton from AoPS

Question is here.

You have two congruent triangles. 17-same angle and- x (SAS)
Using distance formula, the two green lines are of the same length.
\(\left( a-25\right) ^{2}+\left( 20-15\right) ^{2}=a^{2}+20^{2}\)
a = 5

Use another distance formula to get x -- the blue line.
\(\sqrt {\left( 5-17\right) ^{2}+20^{2}}=\sqrt {544}= 4\sqrt {34}\)

Thursday, December 5, 2013

Sum and Product of roots : Vieta -- > Questions from 2010-2011 Mathcounts Super Stretch

Questions: (detailed solutions below)

#1 : What is the sum of the solutions of 6x2 + 5x - 4 = 0 ? Express your answer as a common fraction. 

#2 : A quadratic equation of the form x2 + kx + m = 0 has solutions x = 3 + 2 2  and 3 - 2 2 
What is the value of k + m? 

#3 : What is the sum of the reciprocals of the solutions of 4x2 - 13x + 3 = 0 ? Express your answer as a common fraction. 

Solutions : 

#1:  6x2 + 5x -4 = 0 divided the whole equation by 6 and you have x2 + (5/6) x - 4/6 = 0, which means that the sum of the solutions is - 5/6. 

#2: The two roots are 3 + 2 2  and 3 - 2 2 , which means that -k = 3 + 2 2 + 3 - 2 2 
k = -6;  m = (3 + 2 2 ) (3 - 2 2 ) = 9 - 8 = 1 so m + k = -6 + 1 = -5

Solution I:
4x2 - 13x + 3 = 0; divided the whole equation by 4 and you have  x2 - (13/4)x + 3/4 = 0,
which means that the sum of the two roots, if they are x and y, are 13/4 and their product is 3/4.

1/x + 1/y = (x + y) / xy = 13/4  divided by 3/4 = 13/3

Solution II: Tom shows Rob and Rob shows me how to solve this using another method.

The original equation is 4x2 -13x + 3 = 0 To find the sum and product of the reciprocals, you flip the equation so it becomes 3x2 - 13x + 4 = 0

Using the same way you find the sum of the two roots,the answer is 13/3.

Monday, November 4, 2013

Find the area of the petal, or the football shape.

Find the area of the football shape, or the petal shape.
The below Mathcounts mini presents two methods.

Circle and area revisited from Mathcounts mini

The first question is exactly the same as this one.
Besides the two methods on the videos, you can also use the following methods.

Solution III:
You can also look at this as a Venn Diagram question.
One quarter circle is A and the other is B, and both are congruent. (center at opposite corner vertexes)

The overlapping part is C.

A + B - C = 6^2 so C = A + B - 36 or 18pi - 36

Solution IV:
If you use the area of the rectangle,
which is 6 x 12 minus, the area of the half circle with a radius 6, you get the two white spots that are un-shaded.

Use the area of the square minus that will again give you the answer.
\(6^{2}-\left( 6*12-\dfrac {6^{2}\pi } {2}\right)\)
= 18pi - 36

Similar triangles and triangles that share the same vertexes/or/and trapezoid

Another link from my blog

Similar triangles, dimensional change questions are all over the place so make sure you really
understand them.

Take care and happy problem solving !!

Friday, November 1, 2013

Counting I : Ways to Avoid Over Counting or Under Counting

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

Question #1: How many ways to count a triple of natural numbers whose sum adds up to 12 
if  A. order doesn't matter? B. if order matters?

Solution I:
A. Find a systematic way to solve this type of problem in an organized manner.
We can start with the smallest natural number "1".
(1, 1, 10), (1, 2, 9), (1, 3, 8), (1, 4, 7), (1, 5, 6)  Since (1, 5, 6) is the same as (1, 6, 5) so stop.
(2, 2, 8), (2, 3, 7), (2, 4, 6), (2, 5, 5)
(3, 3, 6), (3, 4, 5)
(4, 4, 4) so total there are 12 ways.

B. Since in this case order matters so if you add up all the possible ways, for example, there are "3" ways to arrange (1, 1, 10) -- 3!/2! = 3
This is similar to "how many ways to arrange the letters 'odd'.
If all letters/numbers are different, you have 3! ways to arrange them; however, in this case, the two numbers 1, 1 are indistinguishable and there are 2! ways to arrange them, thus 3!/2! = 3

There are 3! or 6 ways to arrange triples such as (1, 2, 9).
Add all the possible ways and there are total 55 ways.

Solution II:
B.There is a much easier way to tackle this question, using bars and stars or sticks and stones method.
There are 11 spaces between 12 stones. @__@__@__@__@__@__@__@__@__@__@__@
If you places the two sticks on two of the spaces, you'll split the stones into three groups.
For example, if you have @ | @@@@@ | @@@@@@
The triples are 1, 5, and 6.
If it's @@@@@@@@ | @@ | @@ , the triples are 8, 2, and 2.
So 11C2 = 55 ways

Let me know if you are still confused.  Have fun problem solving !!

Monday, September 9, 2013

Chicken, Rabbit Questions : Algebra Without Using Variables

The problems below are much easier so if you are preparing for the state, try the online timed test here.
Just write down random name to enter the test site is fine. At the end of the test, you'll see the answers for the ones you get wrong. Everyone is welcome to take the test. Thanks a lot and have fun problem solving. 

#1: There are 20 horses and chickens at Old Macdonald's farm. Together there are 58 legs. How many horses and how many chickens?

Solutions I :
#1: Using algebra, you have 
H + C = 20---equation 1 and
4H +2C = 58---equation 2

To get rid of one variable you can times equation 1 by 2 to get rid of Chicken or times 4 to get rid of horses. 
Times 2 and you have 2 H + 2 C = 40 (every term needs to be multiplied by 2)---equation 3
                                  4 H + 2C = 58---equation 2
Using equation 2 - equation 3 and yo get 2H = 18 so H = 9 and from there, solve for C = 11

Solution II: 
Let there be C chickens and (20-C) horses. [Since the sum of the number of chickens and horses is 20. If one has C number, the other has (20 - C)

2C + 4 (20 - C) = 58;  2C + 80 - 4C = 58; -2C = -22; C = 11 
and from there, you get H = 20 - 11 = 9

Solution III:
Without using algebra, you can make all the animals be chickens first. In that case, you'll have 20 x 2 = 40 legs.  Since you have 58 legs, you need to get rid of some chickens and bring in more horses. 

You gain 2 legs by every transaction (-2 + 4 = 2).  (58 - 40) / 2 = 9 so 9 horses and 11 chickens.

Other similar questions to practice (answer key below):  
 #1: Rabbits and ducks -- 30 animals and  86 feet. 
#2: There are 24 three-leg stools and four-leg tables. Together there are 86 legs.
#3: There are 43 bicycles and tricycles and together there are 100 wheels.
#4: There are 33 octopus (8 arms) and sea otters and together they have 188 arms/or for sea otters--legs. How many octopus and how many sea otters?  
#5: There are 18 animals in the barnyard, some are cows and some are chickens. There are total 48 legs. How many chickens and how many cows? 

#1: 13 rabbits and 17 ducks. 
#2: 10 three-leg stools and 14 four-leg tables.
#3: 29 bicycles and 14 tricycles. 
#4: 14 octopus and 19 sea otters.  
#5:  6 cows and 12 chickens.

Monday, June 3, 2013

This Week's Work : Week 15 - for Inquisitive Young Mathletes

Part I work for this week:

See if you can write proof to show the exterior angle of any regular convex polygon is \(\frac{360}{n}\).
I'll include that in my blog for better proof.

Polygons Part I : interior angle, exterior angle, sum of all the interior angles in a polygon, how many diagonals
in a polygon

Polygons Part II : reviews and applicable word problems

Interior angles of polygons from "Math Is Fun"

Exterior angles of polygons from "Math Is Fun"

Supplementary angles

Complementary angles

Get an account from Alcumus and choose focus topics on "Polygon Angles" to practice.
Instant feedback is provided. This is by far the best place to learn problem solving, so make the best use of
these wonderful features.

This week's video on math or science : Moebius Transformations Revealed

Part II work online timed test word problems and link to key in the answers will be sent out through e-mail.

Have fun problem solving !!

Tuesday, May 21, 2013

This Week's Work : Week 13 - for Inquisitive Young Mathletes

First, review target level questions you got wrong last week and practice counting systematically.

Right triangle inscribed in a circle proof from Khan Academy

Review 30-60-90 special right triangle from AOPS and practice getting the two legs/hypotenuse fast.

Review 45-45-90 special right triangle from AOPS

The harder questions are the ones when given the length to the 60 degree of the 30-60-90 special right triangle or
the length to the 90 degree of the 45-45-90 special right triangle so make sure you can get them fast and right.

practice here  (instant feedback)

more practices (instant feedback)

See if you can solve all these questions as they are countdown problems.
(hint : lots of Pythagorean triples or applicable special right triangle ratio concept)

from Regents prep (high school level)

Videos/articles on math for this week:


Wednesday, May 15, 2013

This Week's Work : Week 12 -- for Inquisitive Young Mathletes

Link to the online timed test on questions you mostly got wrong or not fast enough. 
(sent through e-mail)

Common Pythagorean Triples: 
3, 4, 5 and its derivatives 
5, 12, 13 
8, 15, 17 
7, 24, 25 (at least these for SAT I and II) 

9, 40, 41, (the rest for state and Nationals, so we'll learn them later)
11, 60, 61
12, 35, 37
13, 84, 85
20, 21, 29

Shoe string method in finding the area of any polygon

Heron's formula in finding the area of a triangle.

Don't mix up the "s" with the other "S" of finding

the area of an equilateral triangle -- proof and formula (You can also use 30-60-90 special right
triangle to get that.)
the area of a regular hexagon

In Heron's case, "s" stands for half of the perimeter.

Besides, I've noticed most of the questions, when given the sides, are best solved by using Pythagorean triples, especially in sprint round questions, so make sure to actively evaluate the question(s) at hand and use the most efficient strategy.

Here is the link to 2003 chapter #29 that most of you got wrong:

You don't need to use complementary counting for that specific question since it's equal cases either way. Make sure you understand why you need to times 3. (AA_, A_A, and _AA for team A to be chosen two out of three days).

From Mathcounts Mini: Area of irregular polygon

See if you can use shoestring method to get the same answer.
Second half is again on similar triangles, dimensional change and sometimes
Pythagorean triples.

From NOVA : Fractals - Hunting the Hidden Dimension

Wednesday, May 1, 2013

This Week's Work : Week 10 -- for Inquisitive Young Mathletes

Learn or review : How many zeros?

Learn or review : Special Right Trianlges

From Art of Problem Solving of some triangles:

Isosceles and Equilateral Triangles

Isosceles Right Triangles (45-45-90 degree special right triangles)

30-60-90 Degree Special Right Triangles

Math and Science go hand in hand so watch this fascinating video from NOVA just for inspiration. 

From NOVA: What Will the Future Be Like? 

Have fun problem soling and exploring !! 

Tuesday, April 23, 2013

This Week's Work : Week 9 -- for Inquisitive Young Mathletes

Part 1:
See below for links:
They are all related to dimensional change and similar polygons.

Dimensional change questions I 

Dimensional change questions II 
Dimensional change questions III : Similar Triangles 

Par II:
Tangent Segments and Similar Triangles from Mathcounts Mini 

If you have more time, download the extra word problems to see if you can solve them at
reasonable speed and accuracy.

Online timed test and problem of the week will be sent out through e-mail.
Time: 40  minutes without a calculator.

The Monty Hall Problem explained

Tuesday, April 16, 2013

This Week's Work : Week 8 -- for Inquisitive Young Mathletes

Assignment 1:
Using Algebra and Number Sense as Shortcuts from Mathcounts Mini

Watch the video and work on the activity sheet below the video on the same link for more practices.

 Also, review the following:
\(x^{2}-y^{2}=\left( x+y\right) \left( x-y\right)\) \(\left( x+y\right) ^{2}=x^{2}+2xy+y^{2}\) \(\left( x-y\right) ^{2}=x^{2}-2xy+y^{2}\) \(\left( x+y\right) ^{2}-2xy =x^{2}+ y^{2}\)
\(\left( x-y\right) ^{2}+ 2xy =x^{2}+y^{2}\)
\(\left( x+y\right) ^{2}-4xy =\left( x-y\right) ^{2}\)

Assignment 2:

Pascal's Triangle  from Math is Fun

Pascal's Triangle and Its Patterns

Assignment 3:
It'll be sent through e-mail.
Happy problem solving !!

This Week's Work : Week 6 and 7 Review -- for Inquisitive Young Mathletes

Watch Joint Proportion from Art of Problem Solving 

Spend some time pondering on "Work" word problems from Purple Math. 
These are some very standard word problems you'll encounter in competition math.  

Review --
dimensional change and probability links from previous weeks.  

More practices on inverse and direct relation 

From Regents Exam Prep 
Link I 

Link II

New concepts:

Height to the hypotenuse
How many ways to arrange the word "banana"? (with elements repeating)
Probability that two of the 3 friends were born on the same week day.

Question : If you can earn 0, 1, 3, 7 or 10 points with each shot and each person has three chances, how many scores can't be made? 

Percentage increase (don't forget to minus the original 100% or 1) is very different from at what percent will it return to the original size or what is the size compared to the original.

This Week's Work : Week 5 -- for Inquisitive Young Mathletes

Evan (a 5th grader in PA) 's Problem of the Week: 
A man notices a sign in Shop-a-Lot that says: "All prices are marked 25% off today only!" He decides to buy a shirt that costs $65.12 before the discount. He then uses a $16 gift certificate and the clerk applies 12% sales tax. What is the final cost of the shirt after all the steps are applied? Express your answer to the nearest hundredth. 
When there is a discount that is 25% off, the original price is going to be 75% times the original price. Therefore, 75% of $65.12 is $48.84. Next, subtracting $16, we obtain $32.84. Finally, applying 12% sales tax, we get $32.84 x 112% (since original price is added to the sales tax "price") which is $36.7808. This rounded to the nearest hundredth is $36.78.

Assignment 1: 

Painted Cube Problems

Visualization of the Painted Cube Problems 

It's more fun if you dig out Legos or Unit Cubes and just build some cubes (and later rectangular prism) 
and spend some time observing how it works. 

Assignment 2:

Review special right triangle, Pythagorean triples, theorem :

30-60-90 , 45-45-90 special right triangle angle ratios

Dimensional Change

Inscribed and Circumscribed Circle Radius of an Equilateral Triangle

This Week's Work : Week 4 -- for Inquisitive Young Mathleges

First of all, problem of the week from Evan, a 5th grader:
A man notices a sign in Shop-a-Lot that says: "All prices are marked 25% off today only!" He decides to buy a shirt that costs $65.12 before the discount. He then uses a $16 gift certificate and the clerk applies 12% sales tax. What is the final cost of the shirt after all the steps are applied? Express your answer to the nearest hundredth. 

This week, we'll learn two very common sequences : arithmetic and geometric sequences.
There are quite a few similarities between these two types and they are closely linked to ratio, proportion
so just watch the videos and play around/generate a few/ponder on those sequences. I don't expect you to learn them in just one week.

Notes from Regents Exam Prep: Arithmetic and Geometric Sequences and Series

From Mthcounts Mini:

Easier concepts:


Arithmetic sequence/determine the nth term

Arithmetic and geometric sequences

Harder concepts:

Relationship between arithmetic sequences, mean and median

Sequences, series and patterns

From my blog : 
Some special arithmetic sequences and the easier way to find their sum

Write some notes of the most important features of the arithmetic sequence.
The best note will be posted here to share with other students.
Be creative !!

This Week's Work : week 3 -- for Inquisitive Young Mathletes

Assignment 1: 
Watch and learn Simon's Favorite Factoring Trick.

Work on some of the problems as well to check your understanding.

Assignment 2: 
Just learn as much as you can.
We'll keep practicing counting and probability.

Counting Permutations : from Art of Problem Solving

With or Without Replacement : from Art of Problem Solving

Notes on Permutations from the Math Page

Permutations with Some Identical Elements

This Week's Work : Week 2 -- for Inquisitve Young Mathletes

Assignment 1:
Mathcounts Mini related to the "Set" concept
Download the word problems below the video and work on them for this week.

Pascal's Triangle from Math is Fun.

Below is problem of the week, which continues with Evan's problem from last week so read it carefully.
Two players play a game starting with a pile of 26 sticks. The players alternate turns, each taking 1, 2, or 3 sticks on his or her turn.The player who takes the last stick wins.Who has the winning strategy in this game, the first player or the second player? How many sticks he/she needs to take? Why? 

Assignment 2:
Review special right triangles: Notes from my blog

30-60-90 Triangles from Art of Problem Solving

Powers of Pythagorean Triples from Art of Problem Solving

Working together rate  problems from Art of Problem Solving.

Thursday, February 21, 2013

2013 Mathcounts State Prep : Inscribed Circle Radius and Circumscribed Circle Radius of a right triangle

Question: \(\Delta\) ABC is a right triangle and a, b, c are three sides, c being the hypotenuse.
What is a. the radius of the inscribed circle and
b. the radius of the circumscribed circle? 

Solution a :
Area of the right \(\Delta\)ABC =  \(\dfrac {ab} {2}\) = \(\dfrac {\left( a+b+c\right) \times r} {2}\)
r =\(\dfrac {ab} {a+b+c}\)

Solution b: 
In any right triangle, the circumscribed diameter is the same as the hypotenuse, so the circumscribed radius is\(\dfrac {1} {2}\) of the hypotenuse, in this case \(\dfrac {1} {2}\) of c or \(\dfrac {1} {2}\) of \(\overline {AC}\)

Some other observations: 
A. If you only know what the three vertices of the right triangle are on a Cartesian plane, you can use distance formula to get each side length and from there find the radius.

B.In right \(\Delta\)ABC , \(\overline {AC}\) is the hypotenuse.
If you connect B to the median of \(\overline {AC}\), then \(\overline {BD}\) = \(\overline {AD}\) = \(\overline {CD}\) = radius of the circumscribed circle

Sunday, February 10, 2013

2013 Mathcounts State Prep: Counting Problems

Please check out Mathcounts, the best middle school competition program up to the national level.

#1: 2006 Mathcounts state : My three-digit code is 023. Reckha can’t choose a code that is the same as mine in two or more of the three digit-positions, nor that is the same as mine except for switching the positions of two digits (so 320 and 203, for example, are forbidden, but 302 is fine). Reckha can otherwise choose any three-digit code where each digit is in the set {0, 1, 2, ..., 9}. How many codes are available for Reckha?

Do complementary counting. Use total possible ways minus those that are not allowed. 
You can't use two or more of the numbers that are at the same position (given) as 203, which means that you can't have 0 __ 3, __ 23, or 02__.

For each of the __, you can use 10 digits (from 0, 1, 2 ... to 9) so 10 + 10 + 10 = 30.

However, you repeat 023 three times in each case so you need to minus 2 back so not to over count.
30-2 = 28

Also, you can't just switch two digits, which means 320, 203 and 032 are not allowed.  { but 302 and 230 are allowed since you are switching all the digits }

There are 10 x 10 x 10 = 1000 digits total and 1000 - 28 - 3 = 969   The answer

#2: 2011 AMC-8 # 23: How many 4-digit positive integers have four different digits where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

For the integer to be a multiple of 5, there are two cases: 

Case I: The unit digit is 5 :  __ __ __ 5
There are 4 numbers to choose for the thousandth digit [since 5 is the largest digit and you can't have "0" for the leading digit so there are 4 numbers 1, 2, 3, 4 that you can use], 4 numbers to choose for the hundredth
digit (0 and one of the remaining 3 numbers that are not the same number as the one in the thousandth digit) and 3 numbers to choose for the tenth digit (the remaining 3 numbers) so total 4 x 4 x 3 = 48 ways

Case II:  The unit digit is 0: __ __ __ 0
One of the remaining three numbers has to be 5, and for the remaining 2 numbers, there are 4C2 = 6 ways
to choose the 2 numbers from the numbers 1, 2, 3 or 4.
There are 3! arrangements for the three numbers so 6 x 3! = 36

48 + 36 = 84 ways

Wednesday, February 6, 2013

2013 Mathcounts State Prep : Angle Bisect and Trisect Questions

Proof : 
2y = 2x + b (exterior angle = the sum of the other two interior angles)
--- equation I

y = x + a (same reasoning as above)
--- equation II

Plug in the first equation and you have
2y = 2x + 2a = 2x + b

2a = b

Here is the link to the Angle Bisector Theorem, including the proof and one example.

Angle ABC and ACB are both trisected into three congruent angles of x and y respectively. 
If given angle "a" value, find angle c and angle b.  

Solution: 3x + 3y = 180 - a

From there, it's very easy to find the value of x + y
and get angle c, using 180 - (x + y).

Also, once you get 2x + 2y, you can use the same method -- 180 - (2x + 2y) to get angle b

Monday, February 4, 2013

Counting II : Practice Counting Systematically

Counting Coins 
Lots of similar questions appear on Mathcounts tests. Be careful when there are limits, for example, the sum of the coins do not exceed ___ or you have to have at least one for each type, etc...


The Hockey Stick Identity from Art of Problem Solving
Same as "Sticks and Stones", or "Stars and Bars" methods

Applicable question: Mathcounts 2008 Chapter #9--During football season, 25 teams are ranked by three reporters (Alice, Bob and Cecil). Each reporter assigned all 25 integers (1 through 25) when ranking the twenty-five teams. A team earns 25 points for each first-place ranking, 24 points for each second-place ranking, and so on, getting one point for a 25th place ranking. The Hedgehogs earned 27 total points from the three reporters. How many different ways could the three reporters have assigned their rankings for the Hedgehogs? One such way to be included is Alice - 14th place, Bob - 17th place and Cecil - 20th place.

Solution I :
Let's see how it could be ranked for Hedgehogs to get 27 points from the three reporters.
A          B         C
1           1        25     1 way for C to get 25 points and the other two combined to get 2 points
1           2        24
2           1        24    2 ways for C to get 24 points and the other two combined to get 3 points.
1           3        23
3           1        23
2           2        23    3 ways for C to get 23 and the other two combined to get 4 points.
25        1         1     25 ways for C to get 1 point and the other two combined to get 26 points.
1 + 2 + 3 + ...25 = \(\dfrac {25\times 26} {2}=325\)
Solution II:
Use 26C2. Look at this questions as A + B + C = 27 and A, B C are natural numbers. To split the objects into three groups (for Alice, Bob, and Cecil), we must put 2 dividers between the 27 objects. (You can't grant "0" point.) There are 26 places to put the dividers, so 26C2 and the answer is \(\dfrac {26\times 25} {2}=325\)

2013 Mathcounts State Prep: Partition Questions

#24 2001 Mathcounts Sate Sprint Round: The number 4 can be written as a sum of one or more natural numbers in exactly five ways: 4, 3+1, 2 + 1 + 1, 2 +2 and 1 + 1 + 1 + 1; and so 4 is said to have five partitions. What is the number of partitions for the number 7?

#2: Extra: Try partition the number 5 and the number 8. 

#24: You can solve this problem using the same technique as counting coins:

7     6    5    4    3    2    1

1                                           1 way
       1                                    1 way
             1                 1           2 ways ( 5 + 2 or 5 + 1 + 1)
                   1    1                  1 way
                   1           1           2 ways ( 4 + 2 + 1 or 4 + 1 + 1 + 1)
                         2     0   1      1 ways
                         1     2           3 ways ( 3 + 2 + 2, 3 + 2 + 1 + 1 and 3 + 1 + 1 + 1 + 1)
                                3           4 ways (2 + 2 + 2 + 1, 2 + 2 + 3 ones, 2 + 5 ones and 7 ones.)

Total 15 ways.

The partitions of 5 are listed below (There are 7 ways total.):

5   4   3   2   1
1                           1 way
     1                      1 way
          1    1           2 ways  (3 + 2 and 3 + 1 + 1)
                2           3 ways  (2 + 2 + 1, 2 + 1 + 1 + 1 and 1 + 1 + 1 + 1 + 1)

There are 22 ways to partition the number 8.

Thursday, January 17, 2013

2013 Mathcounts State Prep: Harder State Questions

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

2004 Mathcounts State Sprint #19: The points (x, y) represented in this table lie on a straight line. The point (13, q) lies on the same line. What is the value of p + q? Express your answer as a decimal to the nearest tenth. 
#19:  Look at the table and you'll see each time x + 2, y would -3. 
-5 to -14 is (-9), three times (-3) so p = 2 + 3 x 2 = 8
p would = 13 when 2 + 5.5 * 2 = 13 so q = -5  +  (5.5) * (-3) = -21.5
p + q = -13.5

2004 Mathcounts State Sprint #24: The terms x, x + 2, x + 4, ..., x + 2n form an arithmetic sequence, with x an integer. If each term of the sequence is cubed, the sum of the cubes is - 1197. What is the value of n if n > 3?
The common difference in that arithmetic sequence is 2 and the sum of the cubes is -1197, which means that these numbers are all odd numbers. (cubes of odd numbers are odd and the sum of odd terms of odd numbers is odd. )

(-5)3 + (-7)3 + (-9)3 = -1197  However, n is larger than 3 (given) so the sequence will look like this:
 (-9)3+ (-7)3+ (-5)3+ (-3)3+ (-1)3 + (1)3 + (3)3 = -1197
 x = - 9 and x + 2n = 3, plug in and you get -9 + 2n = 3;  n = 6

Thursday, January 10, 2013

2013 Mathcounts Basic Concept Review: Some Common Sums/numbers

These are some common sums that appear on Mathcounts often.

\(1+2+3+\ldots +n=\dfrac {n\left( n+1\right) } {2}\)

 \(2+4+6+\ldots .2n=n\left( n+1\right)\)

\(1+3+5+\ldots .\left( 2n-1\right) =n^{2}\)

The above are all arithmetic sequences.

The sum of any arithmetic sequence is average times terms (how many numbers).
Besides, the mean and median are the same in any arithmetic sequence.
Combining these knowledge, along with distributive rules some times
(case in point, sum of multiples of n, etc...) will expedite the calculation.

The "nth" triangular number is \(\dfrac {n\left( n+1\right) } {2}\)

The sum of the first n triangular numbers is \(\dfrac {n\left( n+1\right) \left( n+2\right) } {6}\).

 \(1^{2}+2^{2}+3^{2}+\ldots +n^{2}\) = \(\dfrac {n\left( n+1\right) \left( 2n+1\right) } {6}\)

\(1^{3}+2^{3}+3^{3}+\ldots +n^{3}=\)\(\left[ \dfrac {n\left( n+1\right) } {2}\right] ^{2}\)