Two AIME Problems

[Ender]

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



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).

[d&q]

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

[Ender]

Yup. Correct for #1

[rabbit]

f(84) = 997

[Ender]

f(84) = 997

Correct. Solution?

[Ender]

Oh, and I forgot the smiley face.

[rabbit]

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

[Ender]

That works =) It's a "find" problem, not a "prove" problem, so you don't have to prove the pattern is consistent.

[rabbit]

My Number Theory class taught me to find answers by looking for patterns, so that's what I did