Basic Mathcounts Knowledge
By Henry Dai
What is Mathcounts?
• Mathcounts is a nationwide math competition held in various areas of
the United States.
• Every year, usually around February, Microsoft hosts the Lake
Washington chapter competition and the Washington state
competition.
• The purpose of Mathcounts is to build student passion in math.
Structure
• Sprint Round – 30 questions , 1 point per question, 40 minutes in
total. Counts for your individual score and team score
• Target Round – 8 rounds, 2 questions per round, 2 points per
questions, and 6 minutes per round. Counts for your individual score
and team score.
• Team Round – 10 questions, 2 points per question, 20 minutes in
total. Counts for your team score.
• Countdown Round – Top students compete in fast-paced problem
solving.
What should I study?
• Mathcounts is based off the middle school math curriculum.
However, your common core knowledge will not be enough to
succeed in Mathcounts
•
•
•
•
Combinatorics
Algebra
Geometry
Number Theory
• Use your resources (AOPS textbook, mathisfun, your coach, etc.)
How to succeed in Mathcounts
• Work Hard
• Easier said then done, studying hard is the key to success in all areas. The
key here is to learn concepts and problem solving from past problems
and books.
• Work Smart
• Take it little by little. I recommend spending about 20 minutes on
weekdays to just review some knowledge. Keep a journal to record
progress and things you’ve learned.
• Use your resources. You have so many resources, it would be a shame not
to use them. AOPS Volume 1, online problems, and your coach our all ways
to learn techniques and tips for Mathcounts success.
Mathcounts prep week 1
• What we will cover
Art of Problem Solving Volume 1:
-Chapter 4 (proportions)
-Chapter 5 (using the integers)
-Intro to Chapter 7 (Special factorizations and clever manipulations)
-Intro to Chapter 24 (Sequences and series)
-Intro to Quadratic Equations, if time allows
Chapter 4: Proportions
• Preview:
Direct relationship – The quotient of two quantities is a constant, when one
increases so this the other, when one decreases the other decreases as well.
Inverse relationship – The product of two quantities is a constant, when one
increases the other decreases.
12 inches = 1 foot
10 cm = 1 meter
5280 feet = 1 mile
3 feet = 1 yard
.
.
.
Some practice problems
• 57. The population of a town increases 25% during 1991. By what
percent must it decrease the following year to return to the population
it was at the beginning of 1991?
• 60. If the ratio of 2x – y to x + y is 2 : 3, find the ration x : y.
• 70. The wages of 3 men for 4 weeks is $108. At the same rate of pay,
how many weeks will 5 men work for $135?
• 82. If p is 50% of q and r is 40% of q, what percent of r is p?
The house worker problem
• Example 4-5. It takes 3 days for 4 people to paint 5 houses. How long
will it take 2 people to paint 6 houses?
• Solution: First we identify that the amount of work and time put into
the job is directly proportional to the completion of the job. Thus, we
can set up an equation.
•
•
ℎ𝑜𝑢𝑠𝑒𝑠
𝑝𝑒𝑜𝑝𝑙𝑒 𝑑𝑎𝑦𝑠
6
5
=
2 𝑑𝑎𝑦𝑠
12
=
5
12
• When we cross-multiply, we get 10d = 72, thus d = 36/5.
The A and B working together problem
• Example 3-14. Pipe A can fill a pool in 5 hours, while pipe B can fill it
in four. How long will it take for the two to fill the pool if both are
operating at the same time?
• Solution: How can we solve this problem using algebra?
• Well, 𝑥
1
4
+
1
5
= 1 where x is the answer.
• The question is, how does this work? ¼ and 1/5 represent is the amount each pipe can fill in one
hour. Thus, ¼+ 1/5 is the amount they can complete in one hour. 1 represents 100% of the job being
1
1
1
1
9
done. Therefore, the number of hours, x, is 1/( + ). To compute this, add + = . Dividing a
4
5
4
5
20
number into one is called taking a reciprocal. We take a reciprocal by flipping the numerator and
denominator, thus the answer is 20/9.
Algebraic sums problem
• 54. When three numbers are added two at a time, the sums are 29,
46, 53. What is the sum of all three numbers?
Chapter 4 Summary
• Know direct and inverse relationships and when to apply them
• Understand the concept of percent and decimals
• Remember conversion factors
• Algebraic equations can be very useful when dealing with proportions
• The house worker problem
• The A and B working together problem
• Algebraic sums problem
Chapter 5: Using the Integers
• Preview:
Divisibility rules (1-11 except 7)
Modular arithmetic
Base numbers
Prime Factorization
Least Common Multiple and Greatest Common Factor
Some Practice Problems
• 84. Find the GCF of 36, 27, and 45.
• 92. Find the value of digit A if the five-digit number 12A3B is divisible
by 4 and 9, and A≠B.
• 94. How many ways can a debt of $69 be paid using only $5 bills and
$2 bills.
• 97. When n is divided by 5, the remainder is 1. What is the remainder
when 3n is divided by 5?
Base Numbers
• Converting to base 10
• Convert 212 base 3 into base 10
• Convert 234 base 5 into base 10
• Convert 123 base 6 into base 10
• Converting from base 10
• Convert 345 into base 4
• Convert 278 into base 3
• Convert 122 into base 2
Chinese Remainder Theorem Problem
• 91. Find the smallest possible integer which when divided by 10
leaves a remainder of 9, when divided by 9 leaves a remainder of 8, by
8 leaves a remainder of 7, etc., down to where, when divided by 2, it
leaves a remainder of 1.
Solution to Chinese Remainder Theorem
Problem
• Hint: What is the Least Common Multiple of 1-10.
• Solution: Wow, seems complicated at first. Whenever I see something
complicated I read the problem repeatedly until I find something very unique.
In this problem, I noticed that the remainder is always 1 smaller than the
divisor, ex. 10-9=8-7=2-1=1. How does that help us? Well, if we find the Least
Common Multiple of numbers 1-10 and subtract 1, it is going to meet the
requirements. You see, if we did not subtract 1, the remainders would be 10,
9, 8 , 7…., 1 which is the same thing as 0. If we subtract 1, the remainder
becomes one smaller than the divisor. Thus, the LCM – 1 = 2520 – 1= 2519.
Primes
1. What is the prime factorization of 320?
2. How many factors does 320 have?
3. What is the LCM of 27, 36, and 45?
4. What is the GCF of 27, 36, and 45?
Chapter 5 Summary
• Know how to find the GCF and LCM of a list of numbers
• Divisibility Rules can be applied in many ways
• Know how to convert from and to base 10.
• Chinese Remainder Theorem problem
• Primes are very important
Chapter 7: Special Factorizations and Clever
Manipulations (intro)
• Preview
• Difference of squares concept
• Sum of Squares concept
Some practice problems
• 138. Given that 98762 = 97535376, 𝑓𝑖𝑛𝑑 98772 .
• 139. What is the sum of the prime factors of 216 − 1?
• 143. Factor 𝑥 12 − 𝑦12 as completely as possible with integral
coefficients and integral exponent.
Chapter 24: Sequences and Series (Intro)
• Preview:
• Arithmetic Sequence
• Geometric Sequence
• Infinite Sequence
• Basic Sum formulas
Fundamental Sum Problem
• 1.What is the sum of the first n counting numbers?
• Solution:
1. Find the number of terms in the sequence
2. Find the average of the terms in the sequence
• The definition of average, or mean, is
𝑠𝑢𝑚 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠
,
# 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠
multiply the answers to the first two steps.
•
𝑛 𝑛+1
2
Note: This formula is extremely important!
therefore we
Some Practice Problems
• 1. What is the sum of the first 10 counting numbers?
• 3. Evaluate 3 + 4 + 5 + 6 …… + 20.
• 5. What is the sum of the first 50 odd numbers?
• 6. Evaluate (-10) + (-9) + (-8) …. + 10 + 11.
Next week
• Preview for next week:
• Slope and y-intercept
• Graphing Linear functions
• The Quadratic Formula 𝑥 =
−𝑏± 𝑏2 −4𝑎𝑐
2𝑎
• Factoring Quadratic
• Vieta’s Formula
• More counting and probability
Homework
• Purchase the Art of Problem Solving Volume 1: the basics textbook
and solution manual. It is listed as $42 on AOPS and may be cheaper
on Amazon
• Get started on the assigned AOPS readings
• Go to http://www.artofproblemsolving.com/ for past AMC 8
problems and solutions.
• Go to mathisfun.com to review some of this week’s topics and to
prepare for next week’s topics
• Have a Happy Thanksgiving!