Peaches Today, Peaches Tomorrow....

On the 4th day there was 1 left over. So on 3rd day there were 4 left over. So on 2nd day there were 10 left over. So on 1st day there were 22.
(iii) If we assume that 1 peach is eaten each day: 11/15, 55 - 1 = 54 5/6, 45 - 1 = 44 3/4, 33 - 1 = 32 1/2, 16 - 1 = 15 3/5, 9 - 1 = 8 1/4, 2 - 1 = 1
(iv) One possibility: 96 47/48 29/31 41/43 26/27 10/11 22/23 11/13 26/27 15/17 10/11 12/13 5/7 11/12 5/7 5/7 1/3 leaving 2 peaches.

We received a large number of correct solutions to problem (ii), many with very clear explanations of the successful strategies:

Frannie from NGHS reasoned as follows:

To end up with 1 peach left, the beginning amount of peaches should be even.
There are many ways of trial and error that you can try, and use ways to work it out.
At first I picked the random number 12 which turns out to be:
*12*
1st day eat 7 leave 5
2nd day eat 3 1/2 leave 1 1/2
3rd day eat 1 3/4
DOESNT WORK! because there aren't enough peaches left.

So then I doubled the number and used 24:
*24*
1st day eat 13 leave 11
2nd day eat 6 1/2 leave 4 1/2
3rd day eat 3 1/4 leave 1 1/4
4th day there should be 1 left but in this there are 1 1/4,
so a 1/4 too many peaches are left.

Then I thought there probably wouldnt be 1/2s etc and I was so close (only a 1/4 away)
so I went down to 22 (nearest even number) and then that was right!
*22*
1st day eat 12 leave 10
2nd day eat 6 leave 4
3rd day eat 3 leave 1
then 4th day would have 1 left!

Ian and Charlie from William Lovell School reasoned differently:

Working back from the end:
There is one left at the end, so you add the one from the third day and then double it because he ate half the peaches plus one:
1 + 1 = 2
2 x 2 = 4
Then you add one from the second day, and then double it:
4 + 1 = 5
5 x 2 = 10
Then you add one more from the first day, and then double it:
10 + 1 = 11
11 x 2 = 22
At the beginning the monkey had 22 peaches.

Peter from Lawrence Sherrif School approached it in a similar way:

22 peaches because
Day 4: 1 peach
Day 3: (1 Peach + 1 Peach) x 2 = 4 Peaches
Day 2: (4 Peach + 1 Peach) x 2 = 10 Peaches
Day 1: (10 Peach + 1 Peach) x 2 = 22 Peaches

Louisa from Nottingham High School for Girls not only showed her stategy, but then also showed that her solution worked:

1+1 = 2
2x2 = 4
4+1 = 5
5x2 = 10
10+1= 11
11x2 = 22

therefore:
22 peaches to start
he eats half (11) + 1...
this leaves 10
he eats half again (5) + 1...
this leaves 4
he eats half (2) + 1...
he is left with 1 peach.

Matt and Fred from Albion Heights School resorted to some algebraic thinking:

The answer to this problem can easily be solved by using one of the simplest rules in algebra - reverse the operations.
We began by writing out the problem:
"X" equals the original amount of peaches,
while "A" represents the number of remaining peaches on day 2,
"B" is the amount on day 3,
and "C" is what remained on the fourth and final day.
(X/2-1) = A
(A/2-1) = B
(B/2-1) = C
If we got "C" by dividing X by 2 and subtracting 1 three times, we must do the opposite to figure out "X". So, that means we must add 1 to "C" and multiply it by 2 three times.
(C+1*2) = B
(B+1*2) = A
(A+1*2) = X

C=1
(1+1*2) = 4
(4+1*2) = 10
and (10+1*2) = 22
Therefore, X=22.

Ben also resorted to some algebraic reasoning:

Taking x as the initial number of peaches, as days go by we end up with
Day 1: x/2 - 1
Day 2:1/2(x/2 - 1) - 1
Day 3:1/2(1/2(x/2 - 1) - 1) - 1
Day 4:1/2(1/2(x/2 - 1) - 1) - 1 = 1

Therefore
1/2(1/2(x/2 - 1) - 1) - 1 = 1
1/2(1/2(x/2 - 1) - 1) = 2
1/2(x/2 - 1) - 1 = 4
1/2(x/2 - 1) = 5
x/2 - 1 = 10
x/2 = 11
x = 22

Well done to you all.