Question

A biologist watches a chameleon. The chameleon catches
flies and rests after each catch. The biologist notices that:

Answer 1

(i) the first fly is caught after a resting period of one minute;
(ii) the resting period before catching the 2mth fly is the same as the
resting period before catching the mth fly and one minute shorter
than the resting period before catching the (2m + 1)th fly;
(iii) when the chameleon stops resting, he catches a fly instantly

Answer 2

(a) How many flies were caught by the chameleon before his first resting
period of 9 minutes?

Answer 3

(b) After how many minutes will the chameleon catch his 98th fly?

Answer 4

(c) How many flies were caught by the chameleon after 1999 minutes
passed?

Answer 5

hint
Let r(m) denote the rest period before the mth catch,t(m) the number of
minutes before the mth catch, and f(n) as the number of flies caught in
n minutes.

Answer 6

what did you get as r(m)?
you might not need t(m)

Answer 7

this is the solution approach also i dont see why can ı post the full solutıon

Answer 8

http://estoyanov.net/files/MATAMATIKA/5993304-A-Collection-of-Problems-Suggested-for-IMO-19592004.pdf
1999 questıon 23

Answer 9

you closed the q, i suppose u have the solution already.. im working on it though...

Answer 10

we have
\[r(2m)=r(m);\quad r(2m)=r(2m+1)-1\qquad m=1,2,3,\ldots\\
m = 2^n\qquad\text{end of each resting period, he catches 2 flies}\\
r(0)=1\implies r(1)=1+1=2\\
r(2)=2\\
r(3)=3\\
r(4)=4\\
r(5)=5\\\vdots
\]
so, we can conclude that by 9min resting period is the 9^th step.
by end of 9min, he catches, \(2^9=512\)
and just before that, he has \(512-2=510\)

Answer 11

@ganeshie8 @electrokid the solution also in the link
wats this base 9 representation