So here are two AIME problems from an AIME test given a while ago (more than 5 years ago). The first one is pretty straightforward -- more of an exercise, actually. The second one is a nice problem.
AIME #5
(http://latex.sidoh.org/?render=%5C%5C%0D%0A%5Clog_8%20a%20%2B%20%5Clog_4%20b%5E2%20%3D%205%0D%0A%5C%5C%0D%0Aand%0D%0A%5C%5C%0D%0A%5Clog_8%20b%20%2B%20%5Clog_4%20a%5E2%20%3D%207%0D%0A%0D%0A)
Find ab.
AIME #7
f(n) is a piecewise function defined as
{ n - 3 if n >= 1000
f(n) = <
{ f(f(n+5)) if n < 1000
Find f(84).
For the first question (not in latex) :
I combined the terms~
simplified~
log8 ab + log4 a²b² = 12
log8 ab + 2log4 ab = 12
change of base~
(log2 ab)/(log2 8 ) + 2(log2 ab)/(log2 4) = 12
log2 ab/3 + 2(log 2ab)/2 = 12
log2 ab/3 + log2 ab = 12
multiplied~
log2 ab + 3log2 ab = 36
4log2 ab = 36
log2 ab = 9
2^9 = ab
ab = 512
Yup. Correct for #1 :)
f(84) = 997
Oh, and I forgot the smiley face. :)
Quote from: Ender on June 17, 2007, 01:38:41 AM
Quote from: rabbit on June 16, 2007, 07:53:58 PM
f(84) = 997
Correct. Solution?
I solved for f(n) 990 <= n <= 1000, since all values < 1000 end up hitting f(99*) somewhere. After doing the first 4 I saw a pattern of f(odd) = 998, f(even) = 997, so f(84) = 997. There's probably a better way I could have done it, but I don't care much, and the calculations took about 45 seconds all told, so I'm happy :)
That works =) It's a "find" problem, not a "prove" problem, so you don't have to prove the pattern is consistent.
My Number Theory class taught me to find answers by looking for patterns, so that's what I did :P