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.

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.