A little monkey had 75 peaches.

Each day, he kept a fraction of his peaches, gave the rest away, and then ate one.
These are the fractions he decided to keep:

1/2,1/4,3/4,3/5,5/6,11/15
In which order did he use the fractions so that he was left with just one peach at the end?

Question

A little monkey had 75 peaches.

Each day, he kept a fraction of his peaches, gave the rest away, and then ate one.
These are the fractions he decided to keep:

1/2,1/4,3/4,3/5,5/6,11/15
In which order did he use the fractions so that he was left with just one peach at the end?

allank · Answer

Hmm...very interesting question. @everyone viewing: Should we use trial and error?

anonymous · Answer

that's basically the way i'm tackling the poblem...^^^

anonymous · Answer

***problem

anonymous · Answer

Well I thought that the denominator of the fraction has be to a factor of 75.

anonymous · Answer

lemme re phase.

allank · Answer

Good point @Hollywood_chrissy , as the fractions are, it seems the monkey didn't use whole peaches sometimes.

anonymous · Answer

the denominator of the fraction must be a factor of the number of peaches at the beginning of that day.

ganpat · Answer

11/15, 5/6, 3/4,  1/2, 3/5 , 1/4... 

i guess, this might be true ...

anonymous · Answer

@Ganpat did you use trial and error??

ganpat · Answer

yup.. at last he has only 1 left.. u can try out urself as well

allank · Answer

Nice one. That works.

ganpat · Answer

thnx :)..

anonymous · Answer

@Ganpat I am having problems.

ganpat · Answer

problem ??

anonymous · Answer

no no, I just was not subtracting away the one that he was eating.

anonymous · Answer

my bad.

ganpat · Answer

cool !!!! good luck..