## Thursday, December 29, 2016

### 2013 Mathcounts School and Chapter Harder Problems

You can now download and discuss with your friends this year's school and chapter problems.
Here is the link to the official Mathcounts website.

Some more challenging problems from this year's Mathcounts school/or chapter problems.

2013 school team #10 : Three concepts are testing here :
Hint:
a. If you get rid of the remainder, the numbers will be evenly divided into 192, so you are looking at
those factors of 192 - 12 = 180

b. To leave a remainder of 12, those factors of 180 that are included in the Set must be smaller than 12, otherwise, you can further divide it.

c. To find the median, make sure to line up the numbers from the smallest to the largest and find the middle numbers. If there are even numbers of factors larger than 12, average the middle two. Otherwise, the middle number is the answer.

$$180=2^{2}\times 3^{2}\times 5$$ so there are (2 + 1) (2 + 1) (1 + 1) = 18 factors

The list on the left side gives you the first 9 and if you times those numbers with "5", you get 9 other factors,which are 5, 10, 20, 15, 18, 60, 45, 90 and 180.

Discard the factors that are smaller or equal to 12 and list all the other factors in order and find the median.

The answer is "36".

2013 Chapter Sprint:
#21: Dimensional change problem : The height of the top pyramid is $$\dfrac {2} {3}$$ of the larger
pyramid so its volume is $$\left( \dfrac {2} {3}\right) ^{3}$$ of the larger pyramid.

$$\left( \dfrac {2} {3}\right) ^{3}\times \dfrac {1} {3}\times \left( \dfrac {36} {4}\right) ^{2}\times 12 =$$ $$96 cm^{3}$$

# 24:  According to the given:   $$xyz=720$$   and   $$2( xy+yz+zx)= 484$$ so
$$( xy+yz+zx )= 242$$

Since x, y and z are all integers, you factor 720 and see if it will come up with the same x, y and z values
for the second condition.

Problem writer(s) are very smart using this number because the numbers "6", "10", "12" would give you
a surface area of 252. (not right)

The three corrrect numbers are "8", "9", and "10" so the answer is $$\sqrt {8^{2}+9^{2}+10^{2}}=$$ $$7\sqrt {5}$$

#25: Geometric probability: Explanations to similar questions and more practices below.

Probability with geometry representations form Aops.

Geometric probability from "Cut the Knots".

#26: This one is similar to 2002 AMC-10B #21, so try that question to get more practices.
2002 AMC-10B #21 link

#27:
$$\dfrac {1} {A}+\dfrac {1} {B}=\dfrac {1} {2}$$
$$\dfrac {1} {B}+\dfrac {1} {C}=\dfrac {1} {3}$$
$$\dfrac {1} {C}+\dfrac {1} {A}=\dfrac {1} {4}$$
Add them up and you have  $$2 * (\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {12}$$

$$(\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {24}$$

$$\dfrac {1} {\dfrac {1} {A}+\dfrac {1} {B}+\dfrac {1} {C}} =$$$$\dfrac {{24}} {13}$$ hours

#28: Hint : the nth triangular number is the sum of the first "n" natural numbers and $$\dfrac {n\left( n+1\right) } {2}$$ is how you use to find the sum.
From there, you should be able to find how many numbers will be evenly divided by "7".

#29 : Circle questions are very tricky so make sure to find more problems to practice accuracy.

#30 :
Solution I:
Read the solution that is provided by Mathcounts.org here.
Solution II:
Case 1 : $$x-1 > 0\rightarrow x > 1$$ Times ( x - 1) on both sides and you have
$$x^{2}-1>8$$ so x > 3 or x < -3 (discard)

Case 2: $$x-1 < 0$$ so $$x < 1$$ $$\rightarrow x^{2}-1 < 8$$ [You need to change the sign since it's negative.]-3 < x < 3. Combined with x < 1 you have the range as -3 < x < 1
The answer is 60%.

2013 Mathcounts Target #7 and 8:

Target question #8 is very similar to 2011 chapter team #10
It just asks differently.
Read the explanations provided on the Mathcounts official website.
They are explained very well.
Let me know if there are other easier ways to tackle those problems.

Hope this is helpful !! Thanks a lot !! Good luck on Mathcounts state.

## Thursday, December 1, 2016

### Mass Points Geometry

Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry
so spend some time understanding it.

Mass Point Geometry : This one is easier to understand with questions that have detailed solutions.

2014-15 Mathcounts handbook Mass Point Geometry Stretch
from page 39 to page 40

(Talking about motivation, yes, there are students already almost finish
this year's Mathcounts' handbook harder problems.)

From Wikipedia

From AoPS

Mass Point Geometry by Tom Rike

Another useful notes

Videos on Mass Point :

Basics

Mass Point Geometry Part I

Mass Point Geometry : Split Masses Part II

Mass Point Geometry : Part III

## Tuesday, November 29, 2016

### Geometry : Harder Chapter Level Quesitons

Question #1 : The area ratio of two equilateral triangles are 4 to 9 and the sum of their perimeter is 30 3 . What is the area of the a. smaller triangle   b. larger triangle?

Solution:
If the area ratio of two similar polygons is 4 to 9, their corresponding line ratio would be  4  to  9
or 2 to 3.[Make sure you know why.]
The perimeter of the two equilateral triangles is 30 3 so the smaller triangle has a perimeter of
2/5 *  30 3 or 12 3. One side is 4 3 . Using the formula of finding the area of an equilateral triangle $$\dfrac{\sqrt3*s^2}{4}$$ , you get the area to be 12 3.

Use the same method to get the area of the larger triangle as 27 3.
You can also use ratio relationship to get the area of the larger triangle by
multiply 12 3 by 9/4.

2007 Mathcounts Chapter Sprint #30: In parallelogram ABCD, AB = 16 cm, DA = 32 cm, and sides AB and DA form a 45-degree interior angle. In isosceles trapezoid WXYZ with WX ≠ YZ, segment WX is the longer parallel side and has length 16 cm, and two interior angles each have a measure of 45 degrees. Trapezoid WXYZ has the same area as parallelogram ABCD. What is the length of segment YZ?

Solution I:
Make sure you know how to get the unknown leg fast. The height of the parallelogram is 82, so the area of the parallelogram is 48 square units. [Check out the special right triangle section here if you can't get the height fast.]

Let YZ of the trapezoid be x and draw the height. Using 45-45-90 degree angle ratio, you'll get the height. (See image above.)
Area of the trapezoid is average of the two bases time height. WX = 16 (given)
$$\dfrac{(16+x)* (16-x)}{4}$$ = 48 ; 256 - x2 = 192 ;  - x2 = - 64;  x = 8 = YZ

Solution II:

Make the y be the height of the trapezoid. YZ = 16 - 2y.  $$\dfrac{(16-2y + 16)}{2}$$ * y = 48
$${(16 -y)* y = 48}$$$$\rightarrow$$ $${16y -y^2 = 48}$$ $$\rightarrow$$ $${y^2 - 16y + 48 = 0}$$ $$\rightarrow$$ $${(y -12)(y -4)=0}$$ $$\rightarrow$$ $${y = 4}$$ or $${y = 12}$$(doesn't work)
YZ = 16 - 2y. Plug in y = 4 and you have  YZ = 8

Solution III: Let the height be y and you have $$\dfrac{(\overline{YZ}+ 16)* y}{2}"$$ = 48 ; ( YZ + 16) * y = 96
When there are some numbers multiply together equal another number, it's a factoring question.
32 * 3 = 96, YZ = 16 (doesn't work)
24 * 4 = 96, YZ =8

## Monday, November 14, 2016

### 2017 Mathcounts State Prep: Some Counting and Probability Questions on Dot Grids

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

#5 1993 Mathcounts National Target : Find the probability that four randomly selected points on the geoboard below will be the vertices of a square? Express your answer as a common fraction.

#5 2004 AMC 10A: A set of three points is chosen randomly from the grid shown. Each three-point (same image as the below question) set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

2007 Mathcounts Chapter Sprint #29 : The points of this 3-by-3 grid are equally spaced
horizontally and vertically. How many different sets of three points of this grid can be the three
vertices of an isosceles triangle?

Solution:
#5 National Target: There are 16C4 = $$\dfrac {16\times 15\times 14\times 13} {4\times 3\times 2\times 1}$$= 1820 ways to select 4 points on the geoboard.

There are 3 x 3 = 9  one by one squares and 2 x 2 = 4 two by two squares and 1 x 1 = 1 three by three squares. (Do you see the pattern?)

There are 4 other squares that have side length of √ 2
and 2 other larger squares that have side length of 5.

9 + 4 + 1 + 4 + 2 = 20 and $$\dfrac {20} {1820}=\dfrac {1} {91}$$

#5: Solution:
AMC-10A: There are 9C3 = $$\dfrac {9\times 8\times 7} {3\times 2\times 1}$$= 84 ways to chose the three dots and 8 of the lines connecting the three dots will form straight lines. (Three verticals, three horizontals and two diagonals.) so
$$\dfrac {8} {84}=\dfrac {2} {21}$$

#29: Solution:
Use the length of the two congruent legs to solve this problem systematically.

There are 16   1 - 1 - $$\sqrt {2}$$    isosceles triangles.
There are 8    $$\sqrt {2}$$  by  $$\sqrt {2}$$ by 2 isosceles triangles. (See that ?)

There are 4     2 - 2 - $$2\sqrt {2}$$  isosceles triangles.
There are 4    $$\sqrt {5}$$  by  $$\sqrt {5}$$ by 2 isosceles triangles.
Finally, there are $$\sqrt {5}$$  by  $$\sqrt {5}$$ by $$\sqrt {2}$$ isosceles triangles.
Add them up and the answer is 36.

## Monday, November 7, 2016

### How Many Zeros?

Problems: (Solutions below.)

#1. 2003 Chapter Team # 7--How many zeros are at the end of (100!)(200!)(300!) when multiplied out?

#2. How many zeros are at the end of 2013!?

#3. How many zeros are at the end of 10! *9!*8!*7!*6!*5!*4!*3!*2!*1!*0!?

#4. What is the unit digit of 10! + 9! + 8! + 7! + 6! + 5! + 4! + 3! + 2! + 1! + 0!?

#5. The number $$3^{4}\times 4^{5}\times 5^{6}$$ written out in full. How many zeros are there are at the end of the number?

#6. How many zeros are at the end of (31!)/(16!*8!*4!*2!*1!)

#7. 2009 National Sprint #18-- What is the largest integer n such that $$3^{n}$$ is a factor of 1×3×5×…×97×99?

Solutions:
#1.For 100!, there are -- 100/5 = 20 , 20/5 = 4 (Stop when the quotient is not divisible by 5 and then add up all the quotients.), or 20 + 4 = 24 zeros.
For 200!, there are 200/5 =40 , 40/5 = 8, and  8/5 = 1, or total 40 + 8 + 1 = 49 zeros.
For 300!, there are 300/5 = 60, 60/5 = 12, and 12/5 = 2, or total 60 + 12 + 2 = 74 zeros.
Add all the quotients and you get 147 zeros.

#2. Use the same method as #1 and the answer is 501 zeros.

#3:Starting at 5!, you have one "0", the same goes with 6!, 7!, 8!, and 9!
10! will give you 2 extra "0"s. Thus total 7 zeros.

#4: Since starting with 5! you have "0" for the unit digit, you only need to check 4! + 3! + 2! + 1! + 0!.
24 + 6 + 2 + 1 + 1 = 34 so the unit digit is 4.

#5: Make sure to prime factorize all the given number sequences, in this case, it's $$3^{4}\times 2^{10}\times 5^{6}$$ after you do that.
2 * 5 = 10 will give you a zero since there are fewer 5s than 2s so the answer is 6 zeros.

#6: You need the same number of 2 and 5 multiple together to get a "0".
31! gives you 30/5 = 6, 6/5 = 1 or 6 + 1 = 7 multiples of 5
31! gives you 10//2 = 15...15/2 = 7...7/2 = 3...3/2 = 1 or 15 + 7 + 3 + 1 = 26 multiples of 2. 16!*8!*4!*2!*1! gives you 4 multiples of 5 and 8 + 4 + 2 + 1 (16!) + 4 + 2 + 1 (8!) + 2 + 1 (4!) + 1(2!)
= 26 multiples of 2.
Thus the multiples of 2s all cancel out, the answer is "0" zeros.

#7: There are 3*1, 3*3, 3*5...3*33 or $$\dfrac {33-1} {2}+1=17$$ multiples of 3.
There are 9*1, 9* 3...9*11 or $$\dfrac {11-1} {2}+1 = 6$$ extra multiples of 3.
There are 27*1, 27*3 or 2 extra multiples of 3.
There is 81*1 or 1 extra multiples of 3.
Add them up and the answer is 26.

## Wednesday, October 26, 2016

### 2014 AMC 8 Results , Problems, Answer key and Detailed Solutions.

E-mail me if you want to join my group lessons for Mathcounts/AMCs prep.
Different groups based on your current level of proficiency.

My students are amazing. They NEVER stop learning and they don't just do math.

Please don't contact me if you just want state/national questions since I'm extremely busy these days with the preps.
You can purchase Mathcounts National tests and other prep materials at Mathcounts store.

You can also use my online blog contents. If you really understand those concepts, I'm sure you can be placed in your state's top 10 in the above average states [Every year I'd received some online students'/parents' e-mails about their (their kids') state results], not the most competitive states, which are crapshoot for even the USAJMO qualifiers, that I know.

Take care and best of luck.

Have fun problem solving and good luck.

2014 AMC-8 problems and solutions from AoPS wiki

#4 : We've been practicing similar problems to #4 so it should be a breeze if you see right away that the prime number "2" is involved. You'll get a virtual bump if you forgot about that again.

#10 : Almost every test has this type of problem, inclusive, exclusive, between, calendar, space, terms, stages... It's very easy to twist the questions in the hope of confusing students, so slow down on this type of question or for the trickier ones, skip it first. You can always go back to it if you have time left after you get the much easier-to-score points.  (such as #12, 13, 18 -- if you were not trolled and others)

#11 : Similar questions appear at AMC-10, Mathcounts.  For harder cases, complementary counting is easier.
This one, block walk is easier.

#12: 1/ 3!

How about if there are 4 celebrities ? What is the probability that all the baby photos match with the celebrities ? only 1 baby photo matches,  only2 baby photos match, 3 baby photos match, or none matches ?

#13: number theory

For sum of odd and /or even, it's equally likely --
odd + odd = even ; even + even = even
odd + even = odd ; even + odd = odd

For product of odd and/or even, it's not equally likely --
odd * even = even ; even * even = even
odd * odd = odd

For product probability questions, complementary counting with total minus the probability of getting odd product (all odds multiply)is much faster.

SAT/ACT has similar type of problems.

#15 : Central angle and inscribed angles --> Don't forget radius is of the same length.
Learn the basics from Regents prep

#17: rate, time and distance could be tricky

Make sure to have the same units (hour, minutes or seconds) and it's a better idea writing down
R*T = D so you align the given infor. better.

Also, sometimes you can use direct/inverse relationship to solve seemingly harder problems in seconds.

Check out the notes from my blog and see for yourself.

#18 : Trolled question. Oh dear !!

1 4 6 4 1     , but it doesn't specify gender number(s) so
(4 + 4)/2^4 is the most likely.

#19 : more interesting painted cube problems --> one cube is completely hidden inside.

Painted cube animation from Fairy Math Tutors

Painted cube review   Use Lego or other plops to help you visualize how it's done.

#20 : Use 3.14 for pi and if you understand what shape is asked, it's not too bad.

#21: You can cross out right away multiples of three or sum of multiples of 3 by first glance.
For example 1345AA, you can cross out right away 3 and 45 (because 4 + 5 = 9, a multiple of 3).
You don't need to keep adding those numbers up. It's easier this way.

#22 : To set up two-digit numbers, you do 10x + y.
To set up three-digit numbers, you do 100x + 10y + z

For those switching digits questions, sometimes faster way is to use random two or three digit numbers, not in this case, though.

#23 : This one is more like a comprehension question. Since it relates to birthday of the month, there are not many two digit primes you need to weight, so 11, 13 and 17. (19 + 17 would exceed any maximum days of the month). From there, read carefully and you should get the answer.

#24 : a more original question --

To maximize the median, which in this case is the average of the 50th and 51st term, you minimized the first 49 terms, so make them all 1s.
Don't forget the 51st term has to be equal or larger than the 50th term.

#25: The figure shown is just a partial highway image. 40 feet is the diameter and the driver's speed is 5 miles per hour, so units are not the same --> trap

I've found most students, when it comes to circular problems, tend to make careless mistakes because there are just too many variables.
Areas, circles, semi-circles, arch, wedge areas, and those Harvey like "think outside the box" fun problems.

Thus, it's a good idea to slow down for those circular questions. Easier said than remembered.

Happy Holiday !!

### ACT worksheets

ACT worksheets

Let me know if you have any other questions. Best and take good care, Mrs. Lin

worksheet :

Similar triangles :

worksheet :

## Monday, October 10, 2016

### 2012 Harder Mathcounts State Target Questions

Check out Mathcounts here -- the best competition math program for middle schoolers up to the
state and national level.

# 6: A semicircle and a circle are placed inside a square with sides of length 4 cm, as shown. The circle is tangent to two adjacent sides of the square and to the semicircle. The diameter of the semicircle is a side of the square. In centimeters, what is the radius of the circle? Express your answer as a decimal to the nearest hundredth. [2012 Mathcounts State Target #6]

#6:  Solution:
Using Pythagorean theory: (2 + r)2 = (4-r)2 + ( 2- r)2
4 + 4r + r2 = 16 - 8r + r2 + 4 - 4r + r2
r2 - 16 r + 16 = 0
Using the quadratic formula You have 8 ± 4√ 3
Only 8 - 4 3 = 1.07 works

There is a Mathcounts Mini #34 on the same question. Check that out !!

The above question looks very similar to this year's AMC-10 B #22, so try that one.
(cover the answer choices so it's more like Mathcounts)

2014 AMC-10 B problem #22

#8: In one roll of four standard, six-sided dice, what is the probability of rolling exactly three different numbers? Express your answer as a common fraction. [2012 Mathcounts State Target #8]

Solution I : Permutation method
If order matters, there are 6 * 5 * 4 * 1 ways to choose the number, 1 being the same number as one of the previous one.
Let's say if you choose 3 1 4 1.

Now for the placement of those 4 numbers on the 4 different dice. There are 4C2 ways to place where the two "1" will
be positioned so the answer is : $$\dfrac {6\times 5\times 4\times 1\times 4C2} {6^{4}}$$ = $$\dfrac{5}{9}$$
Solution II:  Combination method

There are 6C3 = 20 ways to choose the three numbers.

There are 3 ways that the number can be repeated. [For example: If you choose 1, 2, and 3, the fourth number could be 1, 2 or 3.]

There are $$\dfrac {4!} {2!}$$ =12
ways to arrange the chosen 4 numbers.[same method when you arrange AABC]
So the answer is$$\dfrac{20* 3 *12}{6^4}$$ = $$\dfrac{5}{9}$$

## Wednesday, October 5, 2016

### 2017 Mathcounts State Prep: Volume of a Regular Tetrahedron and Its Relationship with the Cube it's Embedded

How to find volume of a tetrahedron (right pyramid) with side length one.

The above link gives you a visual interpretation of the relationship of a regular tetrahedron, its
relationship with the cube that it is embedded and the other kind of tetrahedron (right angle pyramid).

The side of the cube is $$\dfrac {S} {\sqrt {2}}$$ so the volume of the regular embedded tetrahedron is
$$\dfrac {1} {3}\times \left( \dfrac {S} {\sqrt {2}}\right) ^{3}$$=$$\dfrac {1} {3}*\dfrac {s^{3}} {2\sqrt {2}}$$= $$\dfrac {\sqrt {2}S^{3}} {12}$$.

You can also fine the height of the tetrahedron and then $$\dfrac {1} {3}$$*base*space height to get the volume.
Using Pythagorean theory, the hypotenuse S and one leg which is $$\dfrac {2} {3}$$ of the height of the equilateral triangle base, you'll get the space height.

## Tuesday, September 27, 2016

### 2016 AMC-8 Prep

Interesting articles on math for this week:

How to Fall in Love With Math by Manil Suri from the New York Times

The Simpsons' secret formula : it's written by math geeks by Simon Singh from the Guardian

For this week's self studies (part I work):
Review:

2009 #25 : Review using the Harvey method. :D

2008 #25   Don't use the method on the link. Use the much faster method we talked about at our lesson.

2008 #24  Make a chart. Slow down on similar question such as this one.
This type of problem is very easy to make mistake on under or overcounting.
Skip first and definitely slow down and double, triple check.

2007 #25 Read the solution if you don't get the method we talked about at our lesson.
It takes time to develop insights so you need to be patient.
If you understand the method, this question will be easy, right ?
Stay with this question longer.

2007 #24
Aayush's method is faster.
(To get the sum of three digits that is a multiple of 3, you either get rid of 1 or 4 [do you see it jumps by 3, why?] ) , so the answer is 1/2.

2006 #25
I've seen other problems (AMC-10s) using the oddest prime, which is "2", the only prime number that is even.
Thus, make sure you understand this question.

2006#24  Taking out the factor question.
Also, learn "1001 = 7 x 11 x 13"
"23 x 29 = 667"

2005 #25 Venn diagram is your friend.

2005 #24 Working backwards is the way to go.

For part II work :
This week, work on the last 5 problems from AMC-8 year 2010, 2011, 2012, 1999 and 1998.
Here is the link from AoPs.

Sam' original question:
David has a bag of 8 different-colored six-sided dice. Their colors are red, blue, yellow, green, purple, orange, black, and white. What is the probability that David takes out a red die, rolls a 6, then takes out a purple die, and rolls another 6 without replacement?

Solution:
The probability of rolling a 6 on a red die is 1/8 * 1/6 = 1/48. The probability of rolling a purple die and rolling a 6 after that, without replacement, is 1/7 * 1/6 = 1/42. Therefore, to get both events, 1/48 * 1/42 = 1/2016.

Evan's compiled question:
$$\sqrt {18+8\sqrt {2}}=a+b\sqrt {c}$$
a, b and c are positive integers. Find a + b + c.
Solution:
Square both sides and you have $$18+8\sqrt{2}$$ = $$a^2 + 2ab\sqrt {2} + b^2c$$
You can see ab = 4 = 4 x 1 and c = 2
a = 4, b = 1 and c = 2 so the sum is 7.

Sounak's problem:
A rhombus with sides 4 is drawn. It has an angle of 60 degrees. What is length of the longer diagonal?

Solution:
Well first you have to draw the rhombus's height .The resulting triangle will be 30,60, 90 triangle.
We know the hypotenuse is 4 so now we know the rest of the sides are $$2\sqrt {3}$$ and 2.
Now if we draw the diagonal we see that it makes another right angle triangle.
We know the legs of this triangle are the same as the previous lengths so then we know the diagonal is $$4\sqrt {3}$$.

## Friday, September 23, 2016

### First week's review reminders

For 16-17 Mathcounts handbook :

Warm-up 1 : review #3, 7, 9.
Think of ways to get these questions in seconds.

Warm-up 2 : review 17,18,19,20 -- slow down on #18, similar types are very easy to
get wrong the first try

direct and inverse relationship

From AoPS videos :
Instructions for reviewing

2014 School Rounds link here

Sprint :

#18 (trial and error are faster), #20 (loose pennies out first)
#22, #26, #27, #28, #29, #30 -- Read the solutions and make sure to fully understand
what major concept(s) is (are) tested -- these are not hard problems.

Target :

#1 : Calendar questions could be tricky, so slow down a bit.
Check what stage it starts first.

#2 : WOW, this one is a killer, so tedious -- hard to get it right the first time.

# 3 and 4 are in seconds questions.

# 5: One line, two numbers and you are done.
With a calculator, you don't even need to write anything down, right?  :)

#6 : more interesting question

#7 and 8 : standard questions, not hard, just need to be careful and thorough.

Team :

# 2, 3, 5, 6 (very tedious, I suggest skipping first)
8, 9 (more tedious than 8 and 10, so don't work on problems in order), and
#10 (very easy if you've noticed what is actually tested) It's in seconds question.

Agai, scan those questions and only try those that are your weak spots or marked red.

Please check the solution files I'd sent you to learn the better methods.

Whenever you have extra time, use the other links (individually based) to keep learning.

Keep me posted. Have fun at problem solving.

Don't just do math.  :)

Math related video : Making Stuff Faster
which includes "The Travelling Salesman Problem" and a competition between an astrophysicist and a paleontologist on how to move passengers boarding the planes faster

Good luck, from Mrs. Lin

## Wednesday, September 21, 2016

### 2016 AMC-8 prep

Want to join our group for the up-coming AMC 8 test in Nov. ?

The problems are more complex, including many steps, occasionally not going in difficulty order, or/and there are troll/ trap questions, so it's GREAT to deter students to just memorize the formulas,

I do see some brilliant/ inquisitive students who are not good test takers, if you belong to that group, there are other ways to shine.

E-mail me on that. It's really much, much better to just sit back and enjoy the problems.
Check this awesome note from AoPS forum.

You know, the most amazing thing about various competitions is the energy, the pleasure, the spontaneity, the camaraderie and the kindred spirits.

Thanks a lot to those diligent, inquisitive boys and girls for their impromptu, collaborated efforts.

You are one of its kind. :)

2015 unofficial AMC 8 problems and detailed solutions from online whiz kids.

This year's AMC 8 official statistics is rolling in.
Yeah, my boys' and girls' names are there. :)

Move on to the most fun Mathcounts competition and of course, AMC 10 and AIME tests.

My online whiz kids NEVER stop learning because ___ is who they are and it doesn't have to be all math and science related.

E-mail me at thelinscorner@gmail.com if you want to join us who LOVE problem solving (and many other areas equally challenging and engaging.)

Unofficial , official problems, answer key and detailed solutions to 2014 AMC8 test + official statistics.

This year's (2013) AMC-8 results can be viewed here.

2013 AMC-8 problems in pdf format (easier to print out and work on it as a real test)

Try this if your school doesn't offer AMC-8 test.
40 minutes without a calculator.

If you want to use the test to prepare for Mathcounts, cover the multiple choice
options to make the questions harder unless you have to see the choices to answer

2013 AMC-8 problems

2013 AMC-8 answer key

2013 AMC-8 problems with detailed (multiple) solutions

Trickier problems : #18, 21(mainly the wording) and maybe 25 (slightly tricky)

2014 AMC-8 Result Statistics can be viewed here.

2014 AMC-8 problems and solutions from AoPS wiki.

#4 : We've been practicing similar problems to #4 so it should be a breeze if you see right away that the prime number "2" is involved. You'll get a virtual bump if you forgot about that again.

#10 : Almost every test has this type of problem, inclusive, exclusive, between, calendar, space, terms, stages... It's very easy to twist the questions in the hope of confusing students, so slow down on this type of question or for the trickier ones, skip it first. You can always go back to it if you have time left after you get the much easier-to-score points.  (such as #12, 13, 18 -- if you were not trolled and others)

#11 : Similar questions appear at AMC-10, Mathcounts.  For harder cases, complementary counting is easier.
This one, block walk is easier.

#12: 1/ 3!

How about if there are 4 celebrities ? What is the probability that all the baby photos match with the celebrities ? only 1 baby photo matches,  only2 baby photos match, 3 baby photos match, or none matches ?

#13: number theory

For sum of odd and /or even, it's equally likely --
odd + odd = even ; even + even = even
odd + even = odd ; even + odd = odd

For product of odd and/or even, it's not equally likely --
odd * even = even ; even * even = even
odd * odd = odd

For product probability questions, complementary counting with total minus the probability of getting odd product (all odds multiply)is much faster.

SAT/ACT has similar type of problems.

#15 : Central angle and inscribed angles --> Don't forget radius is of the same length.
Learn the basics from Regents prep

#17: rate, time and distance could be tricky

Make sure to have the same units (hour, minutes or seconds) and it's a better idea writing down
R*T = D so you align the given infor. better.

Also, sometimes you can use direct/inverse relationship to solve seemingly harder problems in seconds.

Check out the notes from my blog and see for yourself.

#18 : Trolled question. Oh dear !!

1 4 6 4 1     , but it doesn't specify gender number(s) so
(4 + 4)/2^4 is the most likely.

#19 : more interesting painted cube problems --> one cube is completely hidden inside.

Painted cube animation from Fairy Math Tutors

Painted cube review   Use Lego or other plops to help you visualize how it's done.

#20 : Use 3.14 for pi and if you understand what shape is asked, it's not too bad.

#21: You can cross out right away multiples of three or sum of multiples of 3 by first glance.
For example 1345AA, you can cross out right away 3 and 45 (because 4 + 5 = 9, a multiple of 3).
You don't need to keep adding those numbers up. It's easier this way.

#22 : To set up two-digit numbers, you do 10x + y.
To set up three-digit numbers, you do 100x + 10y + z

For those switching digits questions, sometimes faster way is to use random two or three digit numbers, not in this case, though.

#23 : This one is more like a comprehension question. Since it relates to birthday of the month, there are not many two digit primes you need to weight, so 11, 13 and 17. (19 + 17 would exceed any maximum days of the month). From there, read carefully and you should get the answer.

#24 : a more original question --

To maximize the median, which in this case is the average of the 50th and 51st term, you minimized the first 49 terms, so make them all 1s.
Don't forget the 51st term has to be equal or larger than the 50th term.

#25: The figure shown is just a partial highway image. 40 feet is the diameter and the driver's speed is 5 miles per hour, so units are not the same --> trap

I've found most students, when it comes to circular problems, tend to make careless mistakes because there are just too many variables.
Areas, circles, semi-circles, arch, wedge areas, and those Harvey like "think outside the box" fun problems.

Thus, it's a good idea to slow down for those circular questions. Easier said than remembered.

Happy Holiday !!

## Monday, August 29, 2016

### Mathcounts Strategy: Shoestring (or Shoelace) method of finding the area of any polygon

Check out Mathcounts here, the best competition math program for middle school students.

Shoelace formula from Wikipedia

More on Shoelace

Problems: Solutions below

#1:  Find the area of a quadrilateral polygon given the four end points (3, 5), (11, 4), (7,0) and (9,8) in a Cartesian plane.

#2 2007 Chapter Target Round: A quadrilateral in the plane has vertices at (1,3),  (1,1), (2, 1) and (2006, 2007). What is the area of the quadrilateral?

#3: Find the area of a polygon with coordinates (1, 1), (3, -1),  ( 4, 4), and  (0.3)

#4: What is the number of square units in the area of the pentagon whose vertices are
(1, 1 ), ( 3, -1),  (6, 2), (5, 6), and (2, 5)?

#5: Find the area of a polygon with coordinates ( -6, 0), (0, 5), (3, -2), and (4, 7)

#6: Find the area of a polygon with coordinates (20, 0), (0, 12), (3, 0), (4, -4)

#7: Find the area of a polygon with coordinates (-8, 4), (2, 12), (3, -5), (4, -4)

#8: Find the area of a triangle with coordinate (-8, -4), (-3, 10), (5, 6)

Solution I: Draw a rectangle and use the area of the rectangle minus the four triangles to get the area of the quadrilateral polygon.

Solution II: Using shoestring method. First, plug in the four points. Second, choose one starting point and list the other points in order (either clockwise or counterclockwise)  and at the end, repeat the starting point. The answer is 33 square units.

Use this link to practice finding the area of any irregular polygon. Keep in mind that a lot of the times you don't need to use shoestring method. Be flexible!! Scroll to the middle section.

#2 Answer: 3008 square units

#3: Answer: 10.5 square units

#4: Answer: 22 square units

#5: Answer: 45.5 square units

#6: Answer: 136 square units

#7: Answer: 98 square units

#8: Answer: 66 square units

## Tuesday, August 16, 2016

### Pathfinder

From Mathcounts Mini :

Counting/Paths Along a Grid

From Art of Problem Solving

Counting Paths on a Grid

Math Principles : Paths on a Grid : Two Approaches

Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally?

Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is $$\dfrac {\left( 5+5\right) !} {5! \times 5!}$$ = 252 ways

Question # 2: How many ways can you  move from A to B if you can only move downward and to right?

Solution : There are $$\dfrac {\left( 4+4\right) !} {4!\times 4!}$$ * 2 * $$\dfrac {\left( 4+4\right) !} {4!\times 4!}$$ = 9800 ways from A to B

## Wednesday, July 6, 2016

### Mathcounts Prep : Algebra Manipulation

Note that:

$$\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}$$
$$\left( x-y\right) ^{2}+2xy= x^{2}+y^{2}$$
$$\left(x-y\right) ^{3}+3xy\left( x-y\right) =x^{3}-y^{3}$$
$$\left( x+y\right) ^{3}-3xy\left( x+y\right) =x^3 + y^{3}$$
$$\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}$$

Applicable questions:

Question 1: If x + y = a and xy = b, what is the sum of the reciprocals of x and y?

Solution:
$$\dfrac {1} {x }+\dfrac {1} {y}=\dfrac {x +y} {xy}$$= $$\dfrac {a} {b}$$

Question 2: If $$x^{2}+y^{2}=153$$ and x + y = 15, what is xy?

Solution:
$$\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}$$
$$15^{2}-2xy=153$$$$\rightarrow xy=36$$

Question 3: If $$\left( x+y\right) ^{2}=1024$$ , $$x^{2}+y^{2}$$ = 530 and x > y , what is x - y?

Solution:
$$\left( x+y\right) ^{2}-2xy=x^{2}+y^{2}$$
1024 - 2xy = 530$$\rightarrow 2xy=494$$
$$\left( x-y\right) ^{2}+2xy=x^{2}+y^{2}$$
$$\left( x-y\right) ^{2}=36$$
x - y = 6

Question 4: x + y = 3 and  $$x^{2}+y^{2}=89$$, what is $$x^{3}+y^{3}$$?

Solution:
$$\left( x +y\right) ^{2}-2xy=x^{2}+y^{2}$$
9 - 2xy = 89 $$\rightarrow -2xy=80$$ so xy = -40
$$\left( x+y\right) ^{3}-3xy\left( x+y\right) =27 - 3(-40)* 3 = 27 + 3*40*3 = x ^{3}+y^{3}$$
$$x ^{3}+y^{3}$$= 387

Question #5: If $$x+\dfrac {1} {x}=5$$, what is $$x^{3}+\dfrac {1} {x ^{3}}$$?

Solution:
$$\left( x+\dfrac {1} {x}\right) ^{3}=x^{3}+3x^{2}.\dfrac {1} {x}+3x.\dfrac {1} {x^{2}}+\dfrac {1} {x^{3}}$$
$$5^{3}=x^{3}+3\left( x+\dfrac {1} {x}\right) +\dfrac {1} {x^{3}}$$
125 - 3*5 = $$x^{3}+\dfrac {1} {x ^{3}}$$
The answer is 110.

Question #6 : 2011 Mathcounts state sprint #24 : x + y + z = 7 and $$x^{2}+y^{2}+z^{2}=19$$, what is the arithmetic mean of the three product xy + yz + xz?

Solution:
$$\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}$$
$$7^{2}-2\left( xy+yz+xz\right) =19$$
xy + yz + xz = 15  so their mean is $$\dfrac {15} {3}=5$$

More practice problems (answer key below):

#1:If x + y = 5 and xy = 3, find the value of $$\dfrac {1} {x^{2}}+\dfrac {1} {y^{2}}$$.

#2: If x + y = 3 and $$x^{2}+y^{2}=6$$, what is the value of $$x^{3}+y^{3}$$?

#3: The sum of two numbers is 2. The product of the same two numbers is 5.
Find the sum of the reciprocals of these two numbers, and express it in simplest form.

#4:If $$x-\dfrac {6} {x}$$ = 11, find the value of $$x^{3}-\dfrac {216} {x^{3}}$$?

#5: If $$x+\dfrac {3} {x} = 9$$, find the value of $$x^{3}+\dfrac {27} {x^{3}}$$?

#6:If $$x+\dfrac {1} {x} = 8$$, what is $$x^{3}+\dfrac {1} {x ^{3}}$$?

#1 :$$\dfrac {19} {9}$$
#2: 13.5
#3: $$\dfrac {2} {5}$$
#4: 1529 [ $$11^{3}$$+ 3 x 6 x 11 =1529]
#5: 648   [$$9^{3}$$-3 x 3 x 9 = 648]
#6: 488   [ $$8^{3}$$– 3 x 8 = 488]

## Sunday, July 3, 2016

### Sum of All the Possible Arrangements of Some Numbers

Check out Mathcounts, the best middle school competition math program up to the national level.

Questions to ponder: (detailed solutions below)

#1: Camy made a list of every possible distinct four-digit positive integer that can be formed using each of the digits 1, 2 , 3 and 4 exactly once in each integer. What is the sum of the integers on Camy's list?

#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)

Solutions:
#1:
Solution I:
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.

Solution II:
The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrange
the four numbers.
2.5 (1000 + 100 + 10 + 1) x 24 = 66660

#2:
Solution I:
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976

Solution II:

Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is $$\dfrac{1 + 3+ 5 + 9} {4} = 4.5$$
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements)  + 96 = 4.5 * 11110 * 24 + 96 = 1199976

Other applicable problems: (answers below)

#1: What is the sum of all the four-digit positive integers that can be written with the digits 1, 2, 3, 4 if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30)

#2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B)

#3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer?

#4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer?

#5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer?

#1: 66660
#2: $$\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8$$
4.8 * 11111 =$$\color{red}{53332.8}$$
#3: 133320
#4: 1333272
#5: 93324

## Wednesday, May 25, 2016

### Learn How to Learn by BOGTRO from AoPS forum -- Thanks a bunch !!

I love the following quotes :

Insanity: doing the same thing over and over again and expecting different results.
I previously thought it's from Albert Einstein, but it's not. I love it anyway.

You can practice shooting eight hours a day, but if your technique is wrong, then all you become is very good at shooting the wrong way. Get the fundamentals down and the level of everything you do will rise.”
from Michael Jordan

"When you first start off trying to solve a problem, the first solutions you come up with are very complex, and most people stop there. But if you keep going, and live with the problem and peel more layer of the onion off, you can often times arrive at some very elegant and simple solutions."
- Steve Jobs, 2006

5 ways to Kill your dreams from TED talk

Below, BOGTRO from AoPS has graciously allow me to post his well-thoughtout article on "Learn How to Learn".

I wish more students will read it , and don't just read it once, but many times at different intervals and really internalize the method. It will help you not just with problem solving/competition math, but learning in general.

Learn How to Learn

About a month ago I was PMed by a member, asking for advice as to how to prepare for MATHCOUNTS. I (strangely) get a lot of these types of PMs, but this one was slightly different. Whereas normally I could answer something along the lines of "read Volume 1, do practice tests, profit", this user was complaining that despite having rigorously worked through Volume 1 and CMMS (I still don't know what this is, but it's implied to be a book), he was still scoring only in the low 20s on sprints.
To some extent, I was able to relate. Back in my MATHCOUNTS days, I was doing loads of practice tests, learning new techniques to shave off precious seconds, and even practicing hitting a buzzer quickly. But my results only marginally improved. Gradually I understood that he was facing the exact same problem I was - although we were doing plenty of work, we were doing it in the wrong way.

After some thought, I formulated a long but fairly detailed response. Given that state-national season is rolling around, and with it the usual abundance of "how do I prepare" threads, I'm reproducing it below (with some minor edits). I referenced sprint several times because that was the specific complaint by the user, but obviously you can replace "sprint" with "target" or even "countdown", or any combination thereof.
________________________________________________________________________________________________________________________
You first need to determine why it is that you're getting low scores on sprint.
Are you running out of time?
Making stupid mistakes?
Bad at computation? Or
do you honestly not know how to do the problems?
The former three are rectified with simply a lot of (effective) practice, where I say "effective" because simply blazing through problems, checking your score, and moving on is not going to help you very much. You need to be critically analyzing almost every problem - not just the ones you got wrong. Sure, you don't need to think too hard about your process on #2, but questions that take you longer than you would like, you get wrong, or you do in a "bashy" way need to be reviewed.

Essentially, you should be following something similar to the following process. Of course, this is not something that is going to work for 100% of people. The point here is not that you should be following these guidelines like a bible, but that you need to think about how to get the maximum benefit out of each practice test you take. You may very well find that the below system doesn't work for you (though you should at least give it a chance - it may seem "boring" at first, but after some time you'll be going through it like it's second nature and learning excellent habits along the way), in which case you should come up with an alteration that works for you. If (or more likely when) you choose to develop your own preparation system, keep in mind that the basic elements should be present - rigorous review of problems you got wrong, self-reflection on why you got them wrong, and so on.
• Take any MATHCOUNTS sprint round under contest conditions. It doesn't really matter which one you take, though it should be fairly recent for best results. When you're done, score with a simple checkmark or X system - don't look through the solutions immediately. Make a note of the problems that took you a long time, even if you got them correct.
• Without timing yourself (though you shouldn't spend more than 15 minutes or so), solve the problems that you either got wrong or didn't answer during the test. This will partially tell you if you're getting questions wrong because of time constraints or because you don't know the material.
• At this point you should have 4 separate categories of problems:
• Completely correct - don't worry about these at all. Though there is some benefit to looking these over, they are significantly less important than all the other questions.
• Correct, but took you a long time. Identify why it took you a long time - and if it matters. A problem taking you 2-3 minutes may sound like a killer, but in general if you only have a couple of these questions that's completely fine. Even if there's only one "timesink", you should be looking through alternate solutions to doing these problems. I find that problems that usually cause timesinks are either geometry problems that are semi-direct applications of similar triangles (which are naturally fairly easy to coordinate bash or something similarly slow, but this may take a while) or counting problems where you just listed out the possibilities and counted them up. Unfortunately, many MATHCOUNTS problems have this as their intended solution, so there's not a great deal you can do about those. However, even though there may not be a cleaner solution, minute steps during your bashing may prove important. And in the event that even with optimizations the problem will still take 2-3 minutes, you may want to just skip it altogether even if you know exactly how to do it.
• Incorrect (or blank), but you solved it after the test. These are questions that you know how to do, but you ran out of time doing. Important is to determine how long it took you to solve these questions. If you solved 2 questions in 30 seconds each after the test, clearly that's worse than solving one problem in the second category. These second and third categories are quite similar and should be evaluated against each other (a quite reasonable rule of thumb is to save any counting question that you don't see how to do within ~10 seconds for later).
• Incorrect, and you couldn't solve it after the test. Look up the solution, searching (or even posting) on AoPS if necessary (which you should likely do anyway, as MATHCOUNTS official solutions are often horrendous). If it's a situation where you just forgot something that you really knew, it's easy to pass this off as a fluke and move on. However, this is a grave mistake. Perhaps if it happens once or twice in an otherwise good practice, you can kind of gloss over it. But make a note of it anyway. Whenever you hit two problems in the same general category that you didn't solve (keep your categories broad, but not too broad. "Geometry" is too broad a category, while "trignometric relations in geometric models of algebraic inequalities" is too specific to be helpful. Something like "similar triangles" or "factoring" is a much better type of category), you should immediately stop your practicing and look up the relevant sections in whatever book you have (e.g. Volume 1, or whatever CMMS is, or even just an internet search, etc.). Don't move on until you are confident in that area. By "confident", I don't mean that you can approach these kinds of problems once in a while. I mean that once you identify a question as being in your category, you should be able to solve it relatively quickly at least 75% of the time.
• File away every single problem that you got wrong. Categorize these as either "I solved this afterwards" (include the time it took you to solve it - approximate is fine) or "I didn't solve this afterwards". You will need these later. Take a break - read a book, play some FTW, go outside, play League of Legends, whatever floats your boat. There's not much value in overloading yourself, especially so close to chapter. If you're feeling particularly ambitious, review a chapter on a topic that you have trouble with.There is no point to reviewing topics you already can solve problems in regularly.
• At the end of the week, collect every single problem on your "incorrect problems list". If you're going through a test a day, these shouldn't number more than 50. Do these like you would a test under contest conditions. Compare your results to your incorrect problems paper (how long it took you to solve the problems, and whether you got them correct). The fact that you've seen the problems already should compensate for the fact that you need to work quicker. If you get a problem wrong, do the same process - don't time yourself while solving all of the remaining problems.
• If you got the same problem wrong twice, there are 3 scenarios:
• You got it wrong both times, but finished it after the test both times. This speaks to your (lack of?) time management, something that comes much more naturally with practice. Keep in mind that MATHCOUNTS really only tests a very small amount of concepts (relatively speaking), so working through old problems virtually guarantees that almost all MATHCOUNTS problems will already be more or less familiar to you on test day.
• You couldn't solve it at all the first time, but solved it after the test the second time. This is improvement, so it's perfectly fine.
• You didn't solve it the second time around. This means that you don't understand the concept - back to the books.
• Take all the problems you got correct (during the test) off your "incorrect problems" sheet, and continue to repeat the process from the top.

This may seem like quite a bit of work when typed up here, but in reality it's not. Instead of perpetuating the cycle of "do a practice test, score it, move on, read some books in some disorganized fashion, take another practice test, hope for improvement" (not even necessarily in that order, which is even more problematic), instead we optimize this routine by taking a single practice test a day and making sure that we get everything possible out of it. There are only so many tests, and a frequent complaint is that people have run out of old contests to do. While this may be true, this most likely means that they're not doing the tests properly. A single test with the time taken to reflect, organize, and perform a targeted review is significantly more beneficial than 5 tests taken without a goal in mind.

All in all, this should take at most a little over an hour per day (a little more at the end of the week). You are, of course, welcome to do more, but there's a sort of diminishing returns law past a certain point. Devoting a great deal of time to MATHCOUNTS is going to seem like a serious mistake in hindsight (I was among the most guilty of this), especially if you realize you were spending time incredibly inefficiently. I won't give an exact quote here (simply because I don't remember it and a quick search doesn't turn it up), but one MATHCOUNTS winner (Albert Ni?) said something along the lines of
Quote:
I knew that I wouldn't be the smartest mathlete competing. But I could, quite realistically, be the hardest working one [...]

In MATHCOUNTS, that's all that's required. But quantifying the term "hard work" is necessary - someone who is pushing a boulder from point A 25% of the way to point B is doing a lot of work for very little benefit, while someone who uses a truck to carry the same boulder to point B is doing significantly less work for significantly more benefit. Perhaps as a more accurate analogy, take two people in a shooting contest. As soon as the whistle blows, person A starts shooting haphazardly at his target, hitting it once in a while but constantly having to reload. Person B, on the other hand, takes his time, lines up his shots, and hits the target with deadly accuracy. This is very similar to MATHCOUNTS. Person A is blowing through his material quickly, getting little benefit overall, but naturally with the experience of shooting comes some slight improvement. On the other side of things, person B is taking the time to think about how best to use his limited resources to improve as best he can. Sure, he starts off a bit slower, and at the end of the day he might still have some ammunition left unused, but overall he hits the target more. The first approach is popular because it's very easy to feel like you're doing something - after all, if you're spending 4 hours a day on practice MATHCOUNTS tests, you're outworking everyone else, right? Don't fall into this trap. Line up your shots.