Learn how to do these problems fast and accurately
Learn how to do these problems fast and accurately
The best math program for middle school students
Problems solved correctly: 15, 16, 17, 18, 19, 20, 21, 22
Problems solved correctly: 15, 16, 17, 18, 19, 21, 22, 23
17 wrong confused
20 wrong, time sink
21 wrong
22 right of hands
23 wrong understood solution
24 wrong time sink casework
25 easy casework, right
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.
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).
Adding the two cases gives the total number of admissible drawings: \[ N_{\text{total}} = 360 + 105 = 465 . \]
Exactly one of those drawings is the perimeter \(1\!-\!2\!-\!3\!-\!4\!-\!5\!-\!6\!-\!7\!-\!1\), which yields a convex heptagon.
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.
Q16 → Wrong due to silly mistake
Skipped Q20
Q25 → Not enough time
Was not able to solve; did not understand substitution of
n → n-k in the solution + solution video.
Redo!
Q24 Notes:
Q15, Q17, Q18, Q19, Q21, Q22, Q23 → Correct
Problem. Dennis rolls three fair six-sided dice, obtaining a, b, c ∈ {1,…,6}. Find \[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr]. \]
Try the question first before scrolling down to read the solution.
Solution.
Step 1 — Linearity of expectation.
\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] \;=\; \mathbb{E}[|a-b|]+\mathbb{E}[|b-c|]+\mathbb{E}[|c-a|] \;=\; 3\,\mathbb{E}[|a-b|]. \]Step 2 — Expected absolute difference of two dice.
Let \(X = |a-b|\). Its distribution is
\[ \begin{array}{c|cccccc} d & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \Pr(X=d) & \tfrac{6}{36} & \tfrac{10}{36} & \tfrac{8}{36} & \tfrac{6}{36} & \tfrac{4}{36} & \tfrac{2}{36} \end{array} \] \[ \mathbb{E}[|a-b|] =\frac{1}{36}\bigl(0\cdot6 + 1\cdot10 + 2\cdot8 + 3\cdot6 + 4\cdot4 + 5\cdot2\bigr) =\frac{70}{36} =\frac{35}{18}. \]Step 3 — Final answer.
\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] = 3 \times \frac{35}{18} = \boxed{\tfrac{35}{6}}. \]Ay
5/7
from a 9th grader Ar.
Hello Mrs. Lin, 4/25
This week I reviewed the SAT problems we went over in class. If you could please give me some of those harder SAT problems going forward for homework that would be great. I thought that they were good practice.
I didn’t have a ton of time this week for AMC work, because I have finals for many classes coming up. However, I did do some problems from the 2016 AMC 10 A.
I had some trouble with problems 11, 12 and 9. If we could please go over those that would be great Sorry about the late reflection again.
from a 7th grader A.
2025 chapter test
June 29, 2025
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
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.
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 \( P=\dfrac{40}{720}=\boxed{\tfrac{1}{18}} \).
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 \[ D_3 \;=\; 3!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) \;=\; 6\left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) \;=\; 2. \]
each week a space
SAT 14 #2 wrong , 15 #2
SAT 16, #5, 6, wrong , 17, 3 and 6
2 weeks later
18, #5 and 5
19 vocab. words 11 - 20
SAT practice test 8, later math don't know what to do
2 weeks later
20, #5, 21, #4
more than a month later
22, #4 and 5 , 23, #4, 6, 8, 12
30, #5, 31, #9
32, #4, 5, 33, 2, 9
local math competition : 21-22 meet 1, # 1, 2, 3 right
34, #4, 35, #6 and 8
21-22 math competition meet 2, #1 and 3 , #2 almost
36 #1,2,3 need to work on harder vocabulary words
37 # 6,9
Math 2021-22 meet 2 # 1 and 3
5/6/25 This week for reading I did Crack Sat tests 38 and 39 and I only gave myself 15 minutes for each test to time myself.
For test 38 the questions I got wrong were 1 and 11 and for test 39 the questions I got wrong were 2 and 5.
Although I felt rushed I was able to complete the test while still comprehending everything so that’s good.
For math this week I realized that I struggled a lot on the algebra 2 part of the (local math competition) tests so I decided that I should learn some of the curriculum.
To learn the course I went on Khan academy and did 2 units of the course and I plan on continuing learning the course to help with (local math competition) problems in the future. Thank you.
2010 Mathcounts
Nationals sprint :
22. Side AB of regular
hexagon ABCDEF is extended past B to point X such that AX = 3AB. Given that
each side of the hexagon is 2 units long, what is the length of segment FX?
Express your answer in simplest radical form.
Try this question first before you scroll down for the solution.
2025 Mathcounts state sprint
#22: Let n be a positive integer less than
or equal to 1000. If the last two digits of n are reversed, the resulting
integer is exactly 85 percent of n. What is the sum of the possible values of
n?
Try this question first. Then scroll down for solution.
Let n be a positive integer less than or equal to 1000. If the last two digits of n are reversed, the resulting integer is exactly 85 percent of n. What is the sum of the possible values of n?
Let the original number be:
$$n = 100h + 10t + u$$The number formed by swapping the tens and units digits is:
$$n' = 100h + 10u + t$$According to the problem:
$$n' = \frac{17}{20}n$$So \( n \) has to be divisible by 20 (make sure you know why). This implies:
$$u = 0, \quad t \text{ is even}$$Let:
$$t = 2k, \quad 0 \leq k \leq 4$$Then:
$$n = 100h + 10t = 100h + 20k$$ $$n' = 100h + t = 100h + 2k$$Now compute the difference:
$$n - n' = 18k$$Also, from the given:
$$n - n' = n - \frac{17}{20}n = \frac{3}{20}n$$Equating both expressions:
$$18k = \frac{3}{20}n \Rightarrow n = 120k$$Since \( k \neq 0 \), we get:
$$n = 120k$$Valid values for \( k \in \{1, 2, 3, 4\} \), so the numbers are:
$$120, \quad 240, \quad 360, \quad 480$$Their sum is:
$$120 + 240 + 360 + 480 = 120(1 + 2 + 3 + 4) = 120 \times 10 = \boxed{1200}$$V's record
weekly homework about 30 to 45 minutes
extra videos, links optional
First lesson:
2010 chapter sprint:
Hi, I got a score of 22 and got questions 16, 22, 23, 24,25,27,29, and 30 wrong.
I just guessed these questions because I didn't really find a way to do them.
Second meet:
2011-12 Mathcounts handbook (40 questions total)
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