Peach Rationing

When looking for problems for my students I came across this problem:

Whenever the monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. I wonder how long he could make his peaches last for? Here are his rules: 
 •Each fraction must be in its simplest form and must be less than 1 . 
•The denominator is never the same as the number of peaches left (for example, if there were 45 peaches left, he would not be allowed to keep 44 45 of them). 
Can you start with fewer than 100 peaches and choose fractions so that there is at least one peach left after a week? What is the longest that you can make them last, starting with fewer than 100 peaches? Send us your solutions showing how many peaches you started with and the fractions you used each day.

Link to the problem here: http://nrich.maths.org/2312

So, I decide I would try this problem on my own. So first thing I picked the largest number that was less than 100, but wasn't prime. I determine that prime numbers would not work based on the rules given. If you tried 11, the only denominator you could use is 11 and that is not possible. So I took 99 and kept 32/33 of the peaches. I picked 33 for the denominator because it is the largest factor of 99. To find the numerator, I took 33 and subtracted 1 from it. This gave me 96 and I subtracted 1 from it to get 95. So I did the same thing using 18/19. This result was 90 and after subtracting 1, I got 89. 89 is a prime number and I had to stop here. So I got stuck here.

Instead of trying to start over, I went back to 95 and tried 17/19(one number less on the numerator than 18 I used previously). This gave me 85 minus which is 84 and did not give me a prime number. So from this point on I tried used the greatest factor of the number given and double checked each one before going through with the multiplication. So here were the days I got:
Day 1 99*32/33->96-1->95
Day 2 95*17/19->85-1->84
Day 3 84*41/42->82-1->81
Day 4 81*26/27->78-1->77
Day 5 77*10/11->70-1->69
Day 6 69*22/23->66-1->65
Day 7 65*11/13->55-1->54 (12/13 gave a final answer of 59 which is prime)
Day 8 54*26/27->52-1->51
Day 9 51*15/17->45-1->44 (16/17 gave a final answer of 47 which is prime)
Day 10 44*10/11->40-1->39 (21/22 gave a final answer of 41 which is prime)
Day 11 39*12/13->36-1->35
Day 12 35*4/5->28-1->27 (6/7 gave a final answer of 29 which is prime)
Day 13 27*7/9->21-1->19 (8/9 gave a final answer of 23 which is prime)
Day 14 20*4/5->16-1->15 (9/10 gave a final answer of 17 which is prime)

I continued this pattern at this for Day 15 and 16, but I ran into a problem.

Day 15 15*2/3->10-1->9
Day 16 9*2/3->6-1->5

I wanted to get my final answer down to one. I could not do this with 5 being prime. I knew that I could get my number of peaches down to 4, then I could get a 1 for my final answer. Since I could not do anything with 9, I went back a day to Day 15. I used the fraction 1/3.

Therefore,
Day 15 15*1/3->5-1->4
Day 16 4*1/2->2-1->1

So I was able to get sixteen days with the peaches. This did leave me a couple of questions at the end. The first question was there other numbers that would work? A simple answer would be yes. With being able to pick the fraction, you could come up with many numbers that would still give up more than one peach after a week. The second question I had was could you find a number that lasted more than 16 days. On the website there are no answers, but I think there might be a number that lasts longer. It would take many calculations to do this though.