An, a 7th grader sample student reflection note, or report

from a 7th grader  A. 

2025 chapter test 

total 34 correct

21-30 wrong on sprint

8th was wrong on target

Sent from my iPhone

2024 chapter test

Sprint Round

2 Q19 Algebra Silly mistake Didn't set up the equation correctly Underline important parts of the question

3 Q23 Geometry Didn't know how to do it Learn how to do it

4 Q25 Geometry Didn't know how to do it Learn how to do it

5 Q26 Algebra Didn't know how to do it Learn how to do it

6 Q27 Time, Rate, Distance Didn't know how to do it Learn how to do it

7 Q28 Number Theory Didn't know how to do it Learn how to do it

8 Q29 Probability Silly mistake Didn't count all possible scenarios Learn how to do a faster way to solve these probability problems

9 Q30 Geometry Didn't know how to do it Learn how to do it

10 Target

11 Q5 Alegbra Wasn't sure on how to do it. Learn how to do it

12 Q6 Number Theory Didn't know how to do it Learn how to do it

13 Q8 Number Theory Wasn't sure on how to do it. Learn how to do it

14

15 Total score: 32

Sent from my iPhone

June 3, 2025 

2023 chapter

Sprint

Q20 time understood question but couldn't think of a solution learn how to do it Total score: 33

Q26 radicals calculation error I made a calculation error either adding or converting the radicals double check the calculation

Q27 3-D Geometry didn't know how to do it. learn how to do it

1. Q28 counting didn't know how to do it. learn how to do it
   Q30 probability didn't know how to do it. learn how to do it
   Target
   Q4 geometry had an idea but it did not work learn how to do it
   Q6 probability didn't know how to do it learn how to do it
   Q7 2-D+3-D geometry had an idea but it did not work learn how to do it
   Q8 probability had an idea but it did not work learn how to do it
   Team
   Q4 combinatorics didn't know how to do it learn how to do it
   Q7 counting and geometry didn't know how to do it learn how to do it
   Q8 geometry didn't know how to create a picture that suits this problem learn how to do use imagination to help create a picture that suits this problem
   Q9 probability had an idea but it did not work learn how to do it
   Q10 counting silly mistake didn't read the question right underline important phrases
   
   

2.   June 29, 2025

3. 2021 state sprint
   20/30
   
   Q14 Probability silly mistake I included one but I wasn't supposed to When it says inclusive only count the highest number from 1 to H.
   Q18 Geometry silly mistake Instead of 6^2 I put 4^2 which messed up the answer First always label the given side lengths then find the unknown side lengths
   Q19 Number theory had the right number but my thought process wasn't right For these type of problems first find the number by subtracting the remainder then find the number of divisors then subtract by the amount of the divisors that cannot have that remainder and that is your answer
   Q22 Algebra didn't know how to this problem Learn how to do this problem a fast way and accurate way
   Q24 Algebra didn't know how to this problem Learn how to do this problem a fast way and accurate way
   Q25 Probability Had an idea got close but it didn't work Learn how to do this problem a fast way and accurate way
   Q27 Number theory didn't know how to this problem Learn how to do this problem a fast way and accurate way
   Q28 Counting Had an idea but it didn't work Learn how to do this problem a fast way and accurate way
   Q29 geometry Had an idea got close but it didn't work Learn how to do this problem a fast way and accurate way
   Q30 Algebra didn't know how to this problem Learn how to do this problem a fast way and accurate way

2014 Mathcounts State/National/AMC Prep : More Counting and Probability Problems from Mathcounts Mini / AoPS

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state prep

Counting the Number of Subsets of a Set

Constructive Counting

More Constructive Counting 

Probability and Counting

Probability with Geometry Representations  : Oh dear, the second half part is hilarious.

Probability with Geometry Representations : solution to the second half problem from previous video

Try this one from 1998 AIME #9. It's not too bad.

2015 Mathcounts State/National Prep

Harder concepts from Mathcounts Mini :
Also try the follow-up problems with detailed solutions. Don't do every question. Just the ones you think is hard.

Geometry :
#27 : Area and Volume 

#30: Similar Triangles and Proportional Reasoning

#34: Circles and Right Triangles

#35: Using Similarity to Solve Geometry Problems

#41: Analytic Geometry : Center of Rotation

More questions to practice from my blog post. 

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state/national prep concepts are in the of difficulty.

Counting the Number of Subsets of a Set

Constructive Counting

More Constructive Counting 

Probability and Counting

Probability with Geometry Representations  : Oh dear, the second half part is hilarious.

Probability with Geometry Representations : solution to the second half problem from previous video

Try this one from 1998 AIME #9. It's not too bad.

From Mathcounts Mink : Center of rotation, equal distance from a point to other two points or two lines, angle bisector (last problem on the follow-up worksheet)

Mathcounts Mini : #41 - Analytic Geometry

Problem Solving Strategies : Complementary Counting

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

Download this year's Mathcounts handbook here.

Video to watch on complementary counting from "Art of Problem Solving"

Part 1

Part 2

Question: How many two-digit numbers contain at least one 9?

 At beginning level, kids start to write down all the numbers that contain 9. However, this turns into impossible task if it's a three-digit or four-digit number. So let's try other ways to do it.

9 _ , if 9 is placed as the tens digit, unit digit can be chosen from 0 -9, altogether 10 numbers.
A lot of kids think there are only 9. Be careful. To find how many consecutive terms from number a to b, you do (b-a) +1

_ 9, there will be 9 choices (1-9) this time as the tens digit.(Why?) Zero can't be placed other than unit digit.
So total you have 10 + 9 - 1 (you've counted 99 twice) = 18

This way is better than the first one, but once the numbers become large, you will easily lose track of those double-counting, triple-counting numbers and over count your answers.

Here is a better way to tackle this type of problem: Think about the case of numbers that contain no 9s, and subtracting this value from the total number of two-digit numbers will give you the answer.

There are 99-10 +1 =90 total two digit numbers. There are 8 (the tens digit) x 9 (unit digit) = 72 numbers that contain no 9s. So 90-72 = 18 gives you the answer.

Try this question: 
How many three-digit numbers contain at least one 9?
There are 999 - 100 + 1 or 999 - 99 = 900 three-digit numbers. 
Or 9 x 10 (you can use the digit "0" now" x 10 = 900 three-digit numbers
900 - 8 x 9 x 9 = 252 numbers 

This is called "complementary counting" and there are numerous problems that you can use this strategies to simplify the reasoning. 

Here is another harder problem from 2003 Mathcounts Chapter Sprint Round #29:
Each day, two out of the three teams in a class are randomly selected to participate in a MATHCOUNTS trial competition. What is the probability that Team A is selected on at least two of the next three days? Express your answer as a common fraction.

Solution:  
Use complementary counting. 
If each day two of the team will be chosen, there will be 3C2 = 3 ways to choose the team -- AB, BC, or AC, so 1/3 of the chance that team A won't be chosen and 2/3 of the chance that team A will be chosen. 

Case 1: Team A is not chosen on any of the three days. The probability is (1/3) ³= 1/27.
Case 2: Team A is chosen on one of the three days : The probability is (2/3) times (1/3)² times 3C1 = 6/27 (A - -, - A - or - - A, which is 3C1 = 3 ways)

Total possibilities - none - at least 1 time = at least two times Team A will be chosen 
so the answer is 1 - 1/27 - 6/27 = 20/27

Other applicable problems: (answer key below)

#1: 2006 AMC10 A: How many four-digit positive integers have at least one digit that is a 2 or 3? 

#2: What is the probability that when tossing two dice, at least one dice will come up a "3"? 

#3: If {x,y} is a subset of S={1,2,3,....50}. What is the probability that xy is even?















Answer key: 

#1:  9000 - 7 x 8 x 8 x 8 = 5416

#2:  The probability of the dice not coming up with a "3" is 5/6.
       1 - (5/6)² = 11/36

2. 1 - (25/50) (24/49) -- only odd times odd will give you odd product, the others will all render even product, so the answer is 37/49.



 

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.

11 Qualities Google Looks For In Job Candidates 

In Chinese : 《專題報導》亞裔尖子生 職場難出頭？

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
the question(s) asked.

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. 

Comments :

#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.

Counting Path on a Grid 

This one includes forbidden path finder

 #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 !!

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 4  \(\sqrt {5}\)  by  \(\sqrt {5}\) by \(\sqrt {2}\) isosceles triangles.
Add them up and the answer is 36.  

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

Comments :

#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.

Counting Path on a Grid 

This one includes forbidden path finder

 #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 !!

Show Your Work, Or, How My Math Abilities Started to Decline

Show your work, or, how my math abilities started to decline

I think it's problematic the way schools teach Algebra. Those meaningless show-your-work approaches, without knowing what Algebra is truly about. The overuse of calculators and the piecemeal way of teaching without the unification of the math concepts are detrimental to our children's ability to think critically and logically.

Of course eventually, it would be beneficial to students if they show their work with the much more challenging word problems (harder Mathcounts state team round, counting and probability questions, etc...), but it's totally different from what some schools ask of our capable students.

How do you improve problem solving skills with tons of worksheets by going through 50 to 100 problems all look very much the same? It's called busy work. 

Quote:  "Insanity: doing the same thing over and over again and expecting different results."

Quotes from Richard Feynman, the famous late Nobel-laureate physicist. Feynman relates his cousin's unhappy experience with algebra:

My cousin at that time—who was three years older—was in high school and was having considerable difficulty with his algebra. I was allowed to sit in the corner while the tutor tried to teach my cousin algebra. I said to my cousin then, "What are you trying to do?" I hear him talking about x, you know."Well, you know, 2x + 7 is equal to 15," he said, "and I'm trying to figure out what x is," and I say, "You mean 4." He says, "Yeah, but you did it by arithmetic. You have to do it by algebra."And that's why my cousin was never able to do algebra, because he didn't understand how he was supposed to do it. I learned algebra, fortunately, by—not going to school—by knowing the whole idea was to find out what x was and it didn't make any difference how you did it. There's no such a thing as, you know, do it by arithmetic, or you do it by algebra. It was a false thing that they had invented in school, so that the children who have to study algebra can all pass it. They had invented a set of rules, which if you followed them without thinking, could produce the answer. Subtract 7 from both sides. If you have a multiplier, divide both sides by the multiplier. And so on. A series of steps by which you could get the answer if you didn't understand what you were trying to do.
So I was lucky. I always learnt things by myself.

This Week's Work : Week 51 for Inquisitive Young Mathletes

For this week's work :

Try 2013 Mathworks Math Contest problems from Texas State University

Answer key and statistics can be viewed here. 

I'll send you detailed solutions once we finish all the problems.

Practice more "at least" problems :

2014 AMC-10 B problem 16 

Solution

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state/national prep concepts are in order the of difficulty.

Counting the Number of Subsets of a Set

Constructive Counting

More Constructive Counting 

Probability and Counting

2013 Mathcounts State Prep: Counting and Probability

Question #1: Rolling two dice, what is the probability that the product is a multiple of 3?
Solution I :
As long as one of the two numbers turn up as 3 or multiple of 3 (in this case, a "6"), the product of the two numbers will be a multiple of 3.
There are 6 * 6 = 36 ways to get 2 numbers. Out of the 36 pairs, you can have
3 - 1
3 - 2
3 - 3
3 - 4
3 - 5
3 - 6, 6 ways.
However, there are only 5 ways left if the other die has a 3 since 3 - 3 only counts as once.
1 - 3
2 - 3
4 - 3
5 - 3
6 - 3
Next we look at "at least one number is "6".
6 - 1
6 - 2
6 - 4 (We already used 6 - 3)
6 - 5
6 - 6 so 5 ways.
The other way around, we have
1 - 6
2 - 6
4 - 6
5 - 6  Total 4 ways, so the answer is \(\frac{\Large{( 6 + 5 + 5 + 4 )}}{\Large{36}}\) = \(\frac{\Large{ 20}}{\Large{36}}\) = \(\frac{\Large{5}}{\Large{9}}\).

Solution II: 
The easiest way to solve this problem is to use complementary counting, which is 1 (100% or total possible way) - none of the the multiples of 3 showing up, so 1 - \(\frac{\Large{4}}{\Large{6}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{2}}{\Large{3}}\) * \(\frac{\Large{2}}{\Large{3}}\)=  \(\frac{\Large{5}}{\Large{9}}\)

Question #2: [2002 AMC-12B #16] Juan rolls a fair regular eight-sided die. Then Amal rolls a fair regular six-sided die. What isthe probability that the product of the two rolls is a multiple of 3?
Solution:
Using complementary counting (see solution II of the previous question), 1 - \(\frac{\Large{6}}{\Large{8}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{3}}{\Large{4}}\) * \(\frac{\Large{2}}{\Large{3}}\)=  \(\frac{\Large{1}}{\Large{2}}\)

Problem Solving Strategy: Complementary Counting

#: How many three digit numbers contain the digit "9" at least once?
Solution:
For the "at least" questions: a lot of the time, the easiest way to solve the problems is to use the total number of ways minus the number of ways which do not satisfy the criteria we're looking for.

There are 999 -100 + 1 = 900 three digit numbers.

There are 8 x 9 x 9 = 648 numbers that do not contain the digit "9" at all, so the answer is 900 - 648 = 252

#2: How many three digit numbers have at least two digits that are the same? 
Solution:
Use complementary counting to find how many three digit numbers have no digit that are the same.
9 digit choices for the hundredth digit (no 0), 9 for the tenth digit (0 is allowed) and 8 digit left for the unit digit, so 9 * 9 * 8 = 648
900 (3 digit numbers) - 648 = 252  

#3: How many of the natural numbers from 1 to 600, inclusive, contain the digit "5" at least once?
(The numbers 152 and 553 are two natural numbers that contain the digit 5 at least once, but 430 is not.)
Solution: 
From 1 to 600 inclusive, there are 600 - 1 + 1 = 600 numbers.

___ ___ ___ If we're looking for all the three-digit numbers which do not contain any 5s: In the hundreds place, you can put 0, 1, 2, 3, or 4 (we'll get rid of '000' later). In the tens place, you can put 9 digits, and there are another 9 possible digits for the units digit, so 5 x 9 x 9 = 405.

405 - 1 (to get rid of '000') + 1 (to compensate for not counting the number '600') = 405, which is the total number of numbers which don't use the digit '5'.

600 - 405 =195, which is the answer.

#4: If you toss three coins, what is the probability that at least one coin lands heads up? 
Solution:
There is a \(\dfrac {1} {2}\times \dfrac {1} {2}\times \dfrac {1} {2}=\dfrac {1} {8}\)chance to get all tails = no heads, so if we want at least one head, the answer is 1 - \(\dfrac {1} {8}\)= \(\dfrac {7} {8}\).

Other applicable problems (answers below):
#1: Amy tosses a nickel four times. What is the probability that she gets at least as many heads as tails ?
#2: What is the probability that the product of the top faces on 2 standard die is even when rolled?
#3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms.
#4: 9 fair coins are flipped. What is the probability that at least 4 are heads? 
#5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5?







Answer: #1-\(\dfrac {11} {16}\) ; #2-\(\dfrac {3} {4}\) ; #3- \(\dfrac {7} {8}\) ; #4-\(\dfrac {191} {256}\) #5: 200

Problem Solving Strategy: Probability, Counting, Grid

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 set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

Solution:

#5 National Target: There are 16C4 = (16 x 15 x 14 x 13)/ 4 x 3 x 2 = 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 20/1820 = 1/91

#5 AMC-10A: There are 9C3 = (9 x 8 x 7) / 3 x 2 x 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 

8/84 = 2/21

2016 Mathcounts State Prep : How to Avoid Making Careless Mistakes

Why Can Some Kids Handle Pressure While Others Fall Apart? from the New York Times

Relax! You'll Be More Productive from the New York Times

First, please read Mathcounts "Forms of Answers" carefully for a few times and remind yourself not to write down the unallowable answers.

Stop Making Stupid Mistakes by Richard Rusczyk from Art of Problem Solving

How to Avoid Careless Mathematical Errors from Reddit

Common errors from all my students: 

Number 1 issue with most of the boys/and a few girls who have strong intuition in math is their horrible handwriting since their minds work much faster than their hands can handle. (See!! I'm making excuses for them.)

I call some of my students "second try ___ [Put your name here.] or third try ____ because I don't need to teach them how to solve a problem but the normal pattern is that it has to take them twice or three times to get it finally right. Thus, they need to slow down and organize their thoughts.

Practice your numbers:  4 (it's not 21), 7, 9 (don't dislocate the circle on top), 2 (don't make it looks like a 7), etc...  

35 cents is very different from 0.35 cents.

Unless stated, you write improper fraction as your answer. Make sure to simplify the answer.

Many mistakes happen when it involves fractions, negative numbers, parenthesis, and questions that
have radicals on the denominator and you need to simplify it, so make sure to double check your math.

Some problems takes many steps to reach the final solutions. Some involve lots of data, extreme long strings of information so if those are your weakness, skip them first and go back to them later.

Every point is the same so don't stay with a tedious question too long and panic later not able to finish the last few harder questions which might be much easier to solve arithmetic wise if you know how.

Circle questions trouble many students. Make sure you are aware of what's being tested.
Is it circumference or area, is it diameter or radius, is it linear to area (squared) or vice versa ?
Are the units the same (most frequently appeared errors)?

Don't forget to divide by 2 for the triangle, times 6 if you use area of an equilateral triangle to get the hexagon. For geometry questions or unit conversions, don't do busy work, write down the equivalent equations and cancel like crazy.

Same goes to probability questions, cancel, cancel and cancel those numbers. You don't need to practice mental math multiplication for most of the Mathcounts problems. Think Smart!!

The answer for probability questions can only be 0 to 1, inclusive.

For lots of algebra questions, manipulations are the way to go or number sense. Use the digit clue to help you narrow down the "trial and error" method. 

Make sure you have the terms and space type questions right. Calendar questions, how many numbers (inclusive, exclusive, between, ...), Stage doesn't necessarily start with 1 or 0.
Make sure you don't over-count or under-count.

Mos students got counting questions wrong when it involves limit.

Case in point : How many non-congruent triangles are there if the perimeter is 15? 
Answer: There are 7 of them.  Try it !!

To be continue...

Sep. 4th, 2014 AMC-8 and Mathcounts State/National prep

To prepare for AMC-8 and Mathcounts simultaneously, it's a good idea to delete AMC-8 multiple choice options to make the test more like Mathcounts problems.

There are some tricky questions for AMC-8 test, so if you are at the Mathcounts state level in above average states, you might get better scores on AMC-10 tests than on AMC-8. (Sigh...)

That's what happened to quite a few of my students because their level is way up so they might not as focused as they worked on the more challenging, more interesting questions. Oh dear !!

Review the following very frequently tested concepts. You really don't need a lot of tools (formulas) but deeper understanding, tenacity and the love of thinking outside the box.

Review similar triangles

Review counting and probability

I'll put solutions to some of the "Mass Points" questions soon.

We all love "Mass Points".

Have fun problem solving. Cheers, Mrs. Lin

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 46 - for Inquisitive Young Mathletes

Finally, this year's Mathcounts state problems are clear for discussions.

Check that out here !! We'll talk about those harder problems this week at our group lessons.

Please let me know how you did/what scores you got if you have followed this blog.
So far, there were a few states' top 10, 20, etc.. and two students I met online will go to the Nationals.

Some e-mails make my day and definitely make me more productive.

Thanks a lot in advance.

For this week's work :

If you haven't watched this video, do so this week.

Joint Proportion : In the video, three methods are discussed and as you can see, the last method is much, much
faster so try that to see if you fully understand the relationship now and can implement it on new, seemingly hard
problems.

We'll continue working on some rate, proportion questions (the harder ones) at our lessons besides reviewing and
practicing speed to solve other harder, counting/probability and geometry questions.

New Mathcounts Mini : Recognizing Squares and Solving a Simpler Problem

From TED Talk : The Magic of Fibonacci Numbers presented by Arthur Benjamin

This Week's Work : Week 45 - for Inquisitive Young Mathletes

From Mathcounts Mini : More Constructive Counting

See if you can use "sticks and stones" to solve the following questions (it requires a little twist).

Tossing 3 dice, what is the probability that the sum is 10 ? 

Tossing 4 dice, what is the probability that the sum is 10 ?

Tossing 4 dice, what is the probability that the sum is 18 ? 

If you want to try harder problems, check out this year's AIME II, 1, 3 and 4. (related to what we've learned recently)

This week's new links, working together, direct and inverse proportion, rate, time distance. 

• Section 7.1: Direct Proportion
• Section 7.2: Inverse Proportion
• Section 7.3: Joint Proportion
• Section 7.4: Speed Part 1*
• Section 7.4: Speed Part 2*
• Section 7.4: Rates Part 1*
• Section 7.4: Rates Part 2 - Work Problems*
• Section 7.4: Speed Problem Solving
• 
  

Take care and happy problem solving. 

Cheers, Mrs. Lin 

Similar to 2023 Mathcounts chapter sprint #30, but harder (level 2)

This question is similar to, but more difficult than the 2023 Mathcounts Chapter Sprint #30, which is as follows:

Consider the seven points on the circle shown. If George draws line segments connecting pairs of points so that each point is connected to exactly two other points, what is the probability that the resulting figure is a convex heptagon? Express your answer as a common fraction.

Try the question first before you read the solution down below.

Convex Heptagon Probability (7 points on a circle)

Label the points 1 – 7 clockwise around the circle. Each point must be joined to exactly two others, so the drawing is a 2-regular graph (a disjoint union of cycles).

1 . Count all edge–sets (2-regular graphs) on 7 labelled points

• One 7-cycle.
  Fix the cyclic order: the edges of a 7-cycle correspond to a permutation of the 6 points after 1, with direction ignored. Hence
  \( \dfrac{6!}{2}=360 \) distinct 7-cycles.
• One 3-cycle + one 4-cycle.
  
  1. Choose the 3-cycle: \( \binom{7}{3}=35 \).
  2. Orient the 3-cycle: \( (3-1)!/2 = 1 \) way.
  3. Orient the remaining 4-cycle: \( (4-1)!/2 = 3 \) ways.
  Total  \( 35 \times 1 \times 3 = 105 \) edge–sets.

Adding the two cases gives the total number of admissible drawings: \[ N_{\text{total}} = 360 + 105 = 465 . \]

2 . Count the favourable edge–set

Exactly one of those drawings is the perimeter \(1\!-\!2\!-\!3\!-\!4\!-\!5\!-\!6\!-\!7\!-\!1\), which yields a convex heptagon.

3 . Probability

\[ \boxed{\displaystyle \Pr(\text{convex heptagon})=\frac{1}{465}} \]

---

Checks: the same counting method gives \(70\) total for 6 points (hexagon) and \(3507\) total for 8 points (octagon), agreeing with \(1/70\) and \(1/3507\) respectively.

2015 Mathcounts National sprint #22 level 1 +

 

2015 Mathcounts National sprint #22

If six people randomly sit down at a table with six chairs, and they do not notice that there are name tags marking assigned seats, what is the probability that exactly three of them sit in the seat he or she was assigned? Express your answer as a common fraction.

Try this yourself first (extremely), then scroll down for solutions.

Method 1-Elementary Counting

Choose the three people who sit correctly: \( \binom{6}{3}=20 \).
The remaining three must all sit incorrectly. A quick check (or listing) shows only two ways: \((A\,B\,C)\) or \((A\,C\,B)\).
Total favourable seatings \(=20\times2=40\); total seatings \(=6!\).
Therefore \( P=\dfrac{40}{720}=\boxed{\tfrac{1}{18}} \).

Method 2 – Derangement formula

Pick the three fixed seats: \( \binom{6}{3}=20 \). Derange the other three: \( !3 = 2 \).
Again \( 20\times2 = 40 \) good seatings, so \[ P \;=\; \frac{20\cdot!3}{6!} \;=\; \boxed{\tfrac{1}{18}} . \]

Derange the remaining 3 people. The number of derangements of 3 items is \[ D_3 \;=\; 3!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) \;=\; 6\left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) \;=\; 2. \]