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## Math for fun and glory

### 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 1

2003 AIME II Problem 1. Created by Sal Khan.

## Want to join the conversation?

- So, I know this is a math contest, but what specifically is this contest about?(7 votes)
- This contest is about all of us showing how much effort and hard work we've put into learning and practicing math.(5 votes)

- wouldn't the answer be 672 because you only took the values of a for a=1,2,3 when a could've been a=1,2,3,4,6,12. the problem never had a restriction for the values to repeat(5 votes)
- 1+1=3 because if you take a close look the 1 is a 1 and the 1 is a 3 and the + is a multiplication sign so 1+1=3(4 votes)

- question is p is a prime no and p+14 , p+10 are also prime no I cant understand how to take out p!(3 votes)
- one of fast way to go on is by a guess work. I started with a guess of 3, and it worked.

so, p=3; p+14=17; p+10=13; all three are prime numbers.(5 votes)

- I need help with a MathCounts problem (not for homework, for self interest):

A square is inscribed in a right triangle with legs of 8 units and 15 units. If two of

the vertices of the square lie on the hypotenuse and the other two vertices of the

square lie on the legs of the triangle, what is the length of a side of the square?

Express your answer as a common fraction.(3 votes)- I have tried using multiple Pythagorean Theorems and substituting variables. The book says that the answer is 2040/409. I can't seem to get to the answer(1 vote)

- why aren't they making this a leson like we answer the questions(4 votes)
- Because they are just solutions from AoPS.(1 vote)

- what is the rmo and imo exam?(4 votes)
- IMO stands for the International Mathematical Olympiad, which is the largest,most competitive, and most difficult math competition for high school students around the world. It takes place yearly over a course of two days, where students are given 4.5 hours to solve 3 questions per day. Overall, it's 6 questions and each question is worth 7 points, with a highest score of 42 points (medals/awards are also given out according to scores). None of the questions require calculus to be solved, but rather tests students' creativity on applying their high school maths knowledge to solve the problems.(5 votes)

- How to sign up for the math contest?(2 votes)
- Search up some AMC test-taking sites near you and contact them. It costs approximately $2.25(1 vote)

- I have a question: why doesn't it work to plug in "a = 12/b" into the equation (came out with a quadratic equation problem)? My Work: 12/b * b * 12/b * b = 6 (12b + 24/b) -> 144 = 12(b + 12/b) -> 12 = b + 12/b -> 0 = b squared - 12b + 12 -> (12 +/- square root of(96))/2 -> = 10.9, 1.1(1 vote)
- The result "a = 12/b" comes from a modification of the equation "abc = 6(a+b+c)", so plugging it back into this equation brings you into a circle, without providing new information. Or it should, there is an algebra mistake in your work here.

The LHS is "abc = ab(a+b) = (12/b)b(12/b + b) = 12(12/b + b)".

The RHS is "6(a + b + c) = 6(12/b + b + (12/b + b)) = 6(2(12/b + b) = 12(12/b + b)".

Thus, the equation resolves to true for any value of b. That is, nothing new is learned. :)(2 votes)

- why can't you do ba^2+ab^2=12a+ 12b

then do ba^2-12a+ab^2-12b=0

then a(ab-12) + b (ab-12)=o

then a+b=0 ?(2 votes)- Because ab = 12, so you can't divide by (ab-12), because that's 0, and you can't divide by 0.(1 vote)

- What is AIME and AMC10 and how do you get to do them? What are the requirements?(2 votes)
- It's a math competition that someone can do if he/she is below grade 10.(1 vote)

## Video transcript

I got this problem here
from the 2003 AIME exam. That stands for the American
Invitational Mathematics Exam. And this was actually the
first problem in the exam. The product N of three positive
integers is 6 times their sum, and one of the integers is
the sum of the other two. Find the sum of all
possible values of N. So we have to deal with
three positive integers. So we have three positive
integers right over here. So let's just think about
three positive integers. Let's call them a, b, and c. They're all positive. They're all integers. The product N of these
three positive integers-- so a times b times c is equal
to N-- is 6 times their sum. This is equal to
6 times the sum. Let me do this in another color. So this is their product. So the product N of
three positive integers is 6 times their sum. So this is equal to 6
times the sum of those integers a plus b plus c. And one of the integers is
the sum of the other two. Well, let's just pick c
to be the sum of a and b. It doesn't matter. These are just names, and
we haven't said one of them is larger or less
than the other one. So let's just say
that a plus b is equal to c, that
one of the integers is the sum of the other two.
c is the sum of a plus b. Find the sum of all
possible values of N. So let's just try to do a
little bit of manipulation of the information we
have here, and maybe we can get some relationship
or some constraints on our numbers, and
then we can kind of go through all of
the possibilities. So let's see, we know that
a plus b is equal to c. So we can replace c
everywhere with a plus b. So this expression
right over here becomes ab, which is
just a times b, times c. But instead of c, I'm going to
write an a plus b over here. And then that is equal to
6 times a plus b plus c. And so once again, I'll replace
with the c with an a plus b, and then what does
this simplify to? So on the right-hand side,
we have 6 times a plus b plus a plus b. This is the same thing
as 6 times 2a plus 2b, just added the a's and the b's. And we can factor out a 2. This is the same thing as if
you take out a 2, 6 times 2 is 12 times a plus b. The left-hand side
right over here is still a times b
or ab times a plus b. So ab times a plus b has got to
be equal to 12 times a plus b. So this is pretty
interesting here. We can divide both
sides by a plus b. We know that a plus b cannot be
equal to 0 since all of these numbers have to be
positive numbers. And the reason why I say that
is if it was 0, dividing by 0 would give you an
undefined answer. So if we divide both
sides by a plus b, we get a times b is equal to 12. So all the constraints
that they gave us boiled down to this
right over here. The product of a and
b is equal to 12. And there's only
so many numbers, so many positive integers where
you if you take the product, you get 12. Let's try them out. So let me write
some columns here. Let's say a, b, c. And then we care
about their product, so I'll write that
over here, so abc. So if a is 1, b
is going to be 12. c is the sum of those two, so
c is going to be 13, 1 times 12 times 13. 12 times 12 is 144 plus
another 12 is going to be 156. And just for fun, you
can verify that this is going to be equal
to 6 times their sum. Their sum is 26,
26 times 6 is 156. So this one definitely worked. It definitely worked
for the constraints. And it should because we
boiled down those constraints to a times b need
to be equal to 12. So let's try another one. 2 times 6, their sum is 8. And then if I were to take
the product of all of these, you get 2 times 6
is 12 times 8 is 96. And then we could try 3 and 4. 3 plus 4 is 7. 3 times 4 is 12 times 7. Actually, I should have known
the a times b is always 12, so you just have to multiply
12 times this last column. 12 times 7 is 84. And there aren't any others. You definitely can't go
above 12 because then you would have to deal
with non-integers. You would have to
deal with fractions. You can't do the negative
versions of these because they all have
to be positive integers. So that's it. Those are all of the
possible positive integers where if you take their
products you get 12. We've essentially
just factored 12. So they want us to find the sum
of all possible values of N. Well, these are all
the possible values of N. N was the product
of those integers. So let's just take the sum. 6 plus 6 is 12 plus
4 is 16, 1 plus 5 is 6 plus 9 is 15 plus
8 is 23, 2 plus 1 is 3. So our answer is 336.