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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?
• This contest is about all of us showing how much effort and hard work we've put into learning and practicing math.
• 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
• 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
• question is p is a prime no and p+14 , p+10 are also prime no I cant understand how to take out p!
• 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.
• 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?
• 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
• Because they are just solutions from AoPS.
(1 vote)
• what is the rmo and imo exam?
• 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.
• 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. :)
• 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 ?