Wednesday, October 27, 2010

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?

Sunday, October 24, 2010

The First Three Problems of AMC 12: Explained

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

Problem 1:
Two is 10% of 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?


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?


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!

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 8th term of the Fibonacci sequence.

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.

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.