The Next 3 of AMC 12: Explained

Problem Source: AMC 12

Problem Level 3~4

Problem 4:

A store normally sells windows at $100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How much will they save if they purchase the windows together rather than separately? 

Problem 5:

The average (mean) of 20 numbers is 30, and the average of 30 other numbers is 20. What is the average of all 50 numbers?

Problem 6:

Josh and Mike live 13 miles apart. Yesterday, Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

Solutions:

The First Three Problems of AMC 12: Explained

Problem Level: 3~4
Problem Source: American Math Competitions 12


Problem 1:
Two is 10% of x and 20% of y . What is x-y?


Problem 2:
The equations 2x+7=3 and bx-10=-2  have the same solution. What is the value of b?


Problem 3:
A rectangle with diagonal length x is twice as long as it is wide. What is the area of the rectangle in terms of x?


(answers)

Solving Three Equations

Problem Level :3 (to me)
Problem Source: American Math Competitions 10 #18


A right triangle has perimeter 32 and area 20. What is the length of its hypotenuse?


Solution:


First we need to set up the equations. Just Kidding. We need to first identify the variables: a ,b, c for the side lengths.  

Factorial Manipulation

Problem Level: 2

51!/ 47!

Solution:

For those of you that do not know what the exclamation mark is, it is factorial.

It means to multiply the positive integers until 1.  It is assumed that 0! =1.

Basic Fibonacci Sequences

Problem Level: 1

Find the 8^th term of the Fibonacci sequence.

Solution:

For those that do not know this sequence; here it is;

0,1,1,2,3,5,8,13,21 ...

Starting with zero and to get the next term you add the previous 2 terms.

The eighth term of the sequence would be:

0, 1 ,1 ,2 ,3 ,5, 8, 13?

The first term of this sequence is one so the correct answer would be: 21.

The "zero" term is zero.

Modular Arithmetic

Problem Level: 4
Problem Source: AIME

Find the remainder when 9 x 99 x 999... x999 9s is divided by 1000.

Solution:
This series can be expressed as 9x99x999... mod 1000.

The concept of modular arithmetic is very simple.  Mod means the remainder when the whole thing is divided by something. Say 1 mod 5 is 1.
Note that 9 x 99 x (1000-1) x (10000-1)... mod 1000
9 x 99 x -1 mod 1000^97
This so happens to become -891 mod 1000.
Add 1000 to both sides and you get 109.

The AIME (American Invitational Math Exam) Hard Math Problems can be solved through clever manipulations.

Combined Sequences

Problem Level: 4 

Find the sum of this unique sequence: 1,2,5,8,8,11,14,15,17, 20, 22,23... 121

Hint:

I have decided not to put a solution so I can challenge some people.  Only use the hint when you cannot find a solution.

We see no real pattern in the sequence from one number to another.  Many questions form when two eights show.  This tells us that there is a +0 somewhere. 

After much analysis, most people will give up.

What you need to know is to look at patterns between numbers, not necessarily at consecutive numbers.

But what these people overlook is that this sequence is made up of two sequences:

1,8,15,22... (up to 121) and

2,5,8,11,14,17,20,23... 121

After we have successfully decoded the message, the rest is up to you to finish it.

Arithmetic Sequence

Problem Level: 3

Find the sum of the arithemetic sequence -2, 4, 10 ... 100.

Solution:

The sum of any arithmetic sequence is the number of numbers in the sequence times the average of the sequence. 

The average (arithmetic mean) of the sequence is very easy to find.

(-2 +100 ) /2

=98/2

=49

The average of the sequence is 49.

To find the number of terms can be a bit more challenging.

Because the common difference is 6, try to make the first term 6.

We can do this by adding eight.  The new sequence is  6, 12 ,18 ...108

Dividing each term by 6 leaves us with: 1,2,3... 53.

This shows that there are 53 numbers in this sequence

53 x 49 = 2597

Answer: The sum of the sequence is 2597

We used counting techniques and arithmetic mean to solve this problem.  Most problems require many types of math to solve.

Using Symmetric Sums

Using Symmetric Sums

Problem Level: 4?

Given the equations:

2x+y+z+t+c=15

x+2y+z+t+c=13

x+y+2z+t+c=12

x+y+z+2t+c=11

x+y+z+t+2c=10

Find x+y+z+t+c

Solution:

Now, most people would try to cancel out some variables and manipulate the equation.  You can do that, but that will take a lot of time (and is boring and would make the problem a level 0 problem). Instead, we can find a faster method.

Let S = x+y+z+t+c

Now, we can substitute it in the equations:

x+S=15

y+S=13

z+S=12

t+S=11

c+S=10

These look much nicer than the previous ones.

We can add up all the equations

x+y+z+t+c+5S=61

S = x+y+z+t+c, so 

6S=61

and S=61/6

Answer: x+y+z+t+c=61/6

As you can see this method is not only faster, with skills, you can do it mentally.

Clever Manipulations in Math

Problem Level: 2?

This is a set of numbers.

Starting with ½ , add 1 to both the numerator and denominator and you get a set of numbers.  Multiply these numbers and find the final value.

Solution:

First we should list the ways we could use to solve the problem

We could multiply everything out but that would take a long time.  Instead, we can solve the problem in a faster way.

Notice that the denominators cancel out the numerators:

2 cancels with 2

3 cancels with 3... all the way too 99.

The end result is clearly: 1/100

Clever manipulation, such as canceling out numbers can greatly help a problem becomes easier.

Tricks in the Hinge Theorem

Level 2 Problem ?

Question:

This is a triangle.  Mentally determine if this is a possible triangle.  It is not 180 degrees.

Solution:

Many people who have had a decent course in geometry should find this problem should not be included in the area of "hard math problems".  But then again, many people make mistakes.  Consider the triangle to have side lengths of a,b, and c .The Hinge Theorem states that in any triangle; a+b>c, b+c>a, a+c>b.  When two sides are added up the total should be greater that the third side.

However, in this problem a +b is not >c.  In fact a+b=c which tells us that this is not a possible triangle.  This tells us that this is not a triangle.

This is a good example of this theorem put into action.

Answer: not a triangle

Math Levels of Difficulty

Most math problems have a level of difficulty. These math problems are sorted out by such levels.

Level 1: If you ever see this problem and have studied the subject; it should be a piece of cake for you.

Level 2: Its a bit tricky.

Level 3: Introductory Olympiad Level

Level 4: Intermediate Olympiad Level

Level 5: These are Olympiad level.  If you can not solve them, do not cry.  If you can solve them by the second, in my view, you are a genius.

Level 6*: My undefined Level.  These are insane level questions.

Good Luck