Question 662229
DISCLAIMER:
I do not know how you were expected to solve the problem.
I do not know if there is a much simpler way to solve it.
 
MY IDEAS FOR A SOLUTION:
Colleen found a number of oranges that was one more than a multiple of 3.
Let's call that number {{{3n+1}}} with {{{n}}} being a positive integer.
Colleen took {{{n+1}}}, which was {{{1}}} plus 1/3 of {{{3n}}}, leaving {{{2n}}}.
 
From here on, I get into cumbersome algebra expressions with {{{n}}},
but you could start trying values for {{{n}}} instead, and work with numbers.
 
That {{{3n+1}}} Colleen found, is the number of oranges Bob left.
Bob meant to leave {{{(3n+1)/2}}} for Colleen and {{{(3n+1)/2}}} for Marjorie,
after he took another {{{(3n+1)/2}}} plus {{{2}}} extra for himself.
In other words, the {{{3n+1}}} was 2/3 of the oranges left after Bob took his extra {{{2}}} oranges.
On waking up, Bob had found
{{{(3/2)(3n+1)+2=(9n+3)/2+2=(9n+7)/2}}} oranges.
That was what Marjorie had left, and was 2/3 of what was there
right after Marjorie took the first orange.
On waking up, Marjorie had found 3/2 as many, plus 1, or
{{{(3/2)((9n+7)/2)+1}}} oranges.
{{{(3/2)((9n+7)/2)+1=(27n+21)/4+1=(27n+25)/4}}}
That was {{{N}}}, the total number of oranges the friends had picked:
{{{N=(27n+25)/4}}} <--> {{{4N=27n+25}}}
 
At this point, we can try numbers for {{{n}}},
and check if {{{27n+25}}} is a multiple of 4.
It has to be, because {{{N}}} should be an integer.
We could find that it works for {{{n}}} values of {{{1}}}, {{{5}}}, {{{9}}}, and so on,
while the values for {{{N}}} are {{{13, {{{40}}}, {{{67}}}, and so on.
We may start to see a pattern, and find a general formula for the infinite possible values of {{{N}}}.
 
Alternatively, working from {{{N=(27n+25)/4}}}, we can find that
{{{N=(27n+25)/4}}} --> {{{N=(24n+24+3n+1)/4}}} --> {{{N=6(n+1)+(3n+1)/4}}} --> {{{N=6(n+1)+(3n-3+4)/4}}} --> {{{N=6(n+1)+1+(3n-3)/4}}} --> {{{N=6(n+1)+1+3(n-1)/4}}}
That tells you that {{{n-1}}} must be a multiple of 4 (or zero).
In other words {{{n-1=4k}}} <--> {{{n=4k+1}}} with {{{k}}} a non-negative integer.
Substituting,
{{{N=6(4k+1+1)+1+3*4k/4}}} --> {{{N=6(4k+2)+1+3}}} --> {{{N=24k+12+1+3k}}} --> {{{highlight(N=27k+13)}}} 
 
1. {{{highlight(N=27k+13)}}} with {{{k}}} a non-negative integer.
(that gives you 13, 40, 67, and so on, just keep adding 27)
2. When Colleen departed there were {{{2n=8k+2}}} oranges left.
That could be 2, or 10, or 18, or 26, or ...
3. There are potentially infinity of answers, but I would believe not too many.
It all depends on Colleen's idea of a small number. Collen counted the{{{3n+1=12k+4}}} oranges and was surprised at how small that number was.
No telling what Colleen thinks is a small number.
Maybe {{{12*2+4=28}}} is a small number to Colleen.
Maybe {{{28}}} is a large number to Colleen, and the small number she counted was
{{{12*1+4=16}}} or {{{12*0+4=4}}}.
Maybe Colleen expected more oranges, and finding
{{{12*10+4=124}}}, she thought that was a small number.
4. I'd say Collen counted {{{12*2+4=28}}}, for {{{k=2}}},
which makes {{{N=27*2+13}}} --> {{{highlight(N=67)}}}
5. Colleen took 1 orange first, to make the rest be 3 equal shares.
When, after that, Colleen took 1 of those 3 equal shares, she left 2 equal shares,
so the number left should have been an even number.
It could not have been 5.