It's a New Day
by t0rajir0u, Feb 23, 2007, 5:55 AM
So let's do an AMC problem.
AMC 12 2007B, Problem 24: How many pairs of positive integers
are there such that 
and
Is an integer?
Solution: Writing as one fraction, we have
Is an integer. In order for this expression to be integer, the denominator must evenly divide the numerator. In particular,
Moreover,
But since it follows that 
. So 
.
Case:
. Simplifying, our expression is 
. In order for this expression to be integer, we must have
In fact, all of these give us integer values.
Case:
. Simplifying, our expression is 
. Then we must have
But since this cannot occur. Hence we have a total of 
solutions.
As long as you aren't intimidated by the number of this problem, it isn't as hard as it should've been. Once the connection was made to divisibility some very simple modular logic solved the problem nearly instantly. Then it was just a matter of keeping control of your cases.
Practice Problem: How many integer scores are possible on the AMC?
AMC 12 2007B, Problem 24: How many pairs of positive integers
are there such that 
and
Is an integer?
Solution: Writing as one fraction, we have
Is an integer. In order for this expression to be integer, the denominator must evenly divide the numerator. In particular,
Moreover,
But since
it follows that 
. So 
.Case:
. Simplifying, our expression is 
. In order for this expression to be integer, we must have
In fact, all of these give us integer values.
Case:
. Simplifying, our expression is 
. Then we must have
But since
this cannot occur. Hence we have a total of 
solutions.As long as you aren't intimidated by the number of this problem, it isn't as hard as it should've been. Once the connection was made to divisibility some very simple modular logic solved the problem nearly instantly. Then it was just a matter of keeping control of your cases.
Practice Problem: How many integer scores are possible on the AMC?
April 2009
March 2009