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### Course: Math for fun and glory > Unit 4

Lesson 1: 2003 AIME- 2003 AIME II problem 1
- 2003 AIME II problem 3
- Sum of factors of 27000
- Sum of factors 2
- 2003 AIME II problem 4 (part 1)
- 2003 AIME II problem 4 (part 2)
- 2003 AIME II problem 5
- 2003 AIME II problem 5 minor correction
- Area circumradius formula proof
- 2003 AIME II problem 6
- 2003 AIME II problem 7
- 2003 AIME II problem 8
- Sum of polynomial roots (proof)
- Sum of squares of polynomial roots
- 2003 AIME II problem 9
- 2003 AIME II problem 10
- Trig challenge problem: area of a triangle
- 2003 AIME II problem 12
- 2003 AIME II problem 13
- Trig challenge problem: area of a hexagon
- 2003 AIME II problem 15 (part 1)
- 2003 AIME II problem 15 (part 2)
- 2003 AIME II problem 15 (part 3)

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# 2003 AIME II problem 3

2003 AIME II Problem 3. Created by Sal Khan.

## Want to join the conversation?

- Wow its amazing how simple these questions are! I'm watching these videos because I plan to register my high school for state math league competitions in the fall (so I'm sort of preparing:/), and I always thought these were impossible questions(sort of from experience of being the worst math league team member in middle school). But Mr. Khan makes these soooo easy! Does anyone have any advice for me though?(7 votes)
- Make sure you keep practicing. These skills, such as identifying patterns and being able to "see" potential solutions, easily fade with time. Keep it up! :)(14 votes)

- where`s 2003 AIME II Problem 2(8 votes)
- They probably just decided not to make a video on it. Maybe the problem was "too hard" for them to make a video on. Maybe Sal Khan has just not gotten time to work on the problem in his work office--because he does the videos, he needs to be able to demonstrate how to solve the problem.

Just a side note, the 2003 AIME II problems 1 and 3 seem quite easy for the AIME. I solved each one in two or three minutes, and most AIME problems are expected to take around 10 to 12 minutes to solve...(1 vote)

- for what grade level are these AIME question..?(4 votes)
- There's no set grade level, it depends on your math/problem-solving abilities.

As for what grade you're allowed to take the test, yes, you have to be in 12th-grade or under to officially take the AIME.(1 vote)

- What if C can be immediately followed by A, but other restrictions remain the same?(3 votes)
- Why does multiplying these possibiities gives amount of seven letter good words that can be written?(1 vote)
- This kind of counting is called constructive counting. It is based on the multiplication rule of probability (or whatever it's called).

http://algebralab.org/lessons/lesson.aspx?file=Algebra_ProbabilityMultiplicationRule.xml(1 vote)

- I have no idea what's going on(0 votes)
- How can you tell if B is larger Than C?(0 votes)
- I think you might be a little confused about the problem here. The letters "A", "B", and "C" here are not being used as variables. They're just normal letters.(1 vote)

- This took me maybe 10 seconds to solve; for what grade level is this question?(0 votes)

## Video transcript

Define a "good word" as
a sequence of letters that consists only of
the letters A, B, and C. Some of these letters may
not appear in the sequence. And in which A is never
immediately followed by B, B is never immediately
followed by C, and C is never
immediately followed by A. How many seven letter
"good words" are there? So let's just think
about this a little bit. So there's letters with
just A's, B's, and C's. And then it could be all
A's, all B's, all C's because some letters
might not appear. And A is never
immediately followed by B. So A can only be followed
by another A or another C. B is never immediately
followed by C, which means that B could only be
followed by an A or another B. And C is never
immediately followed by A. So C could only be followed
by another C or a B. So how many seven letter
"good words" are there? So let's just think
about the places. We have seven letters, so
1, 2, 3, 4, 5, 6, 7 letters. Now, there's no constraints
on this first letter since it's not
following anything. So it could be an A
B, or a C. So there's three possibilities
for this first letter. Now, there's three possibilities
for this first letter. But no matter what letter this
is, how many possibilities are there for this second
letter over here? Well, if this was an
A, the second letter could only be an A
or a C because it can't be followed by
a B. If this was a B, the second letter could
only be a B or an A because it can't be followed
by a C. If this was a C, the second letter could
only be a B, or a C. So no matter what letter
this first letter is, the second letter can only
have two possibilities. There could only be
two possibilities. Another way to
think about it is, there's one letter, no
matter what letter this is, there's one letter that's
being ruled out here. So it could only be
two possibilities. Well, the same thing is here. We're going to stick
some letter here. And no matter what letter
there is over here, it's going to rule out
one possibility over here. So we're going to have
only two possible letters that we can put here, no
matter what letter is there. And use the same logic, only
two possibilities there, only two possibilities there,
only two possibilities there, and then only two
possibilities there. So how many total
possibilities do we have? Well, 3 times 2 times 2
times 2 times 2 times 2. This is 1, 2, 3, 4, 5, 6 2's. So this is equal to 3
times 2 to the sixth power, which is 3 times 2 to
the six is 32 times 2 is 64, which is equal to
180 plus 12 is equal to 192. There's 192 possible
good seven letter "good words," where good
words is defined above.