Martin Management

Logic Level 5

$\color{#20A900}{Marloutte}$ is a clever trader , very smart and intelligent. Once, he ordered some apples for trading. The contractor sent his merchant to deliver the apples.

The contractor lived in $\color{#3D99F6}{Mersaills}$ , while $\color{#20A900}{Marloutte}$ lived in $\color{limegreen}{Mosaques}$ . The distance between two cities was $16000$ decametres or $160$ kilometers.

The name of the merchant was $\color{#D61F06}{Martin}$ . His horse, $\color{#69047E}{Mineraques}$ , could carry a maximum of $q$ apples at a time. $\color{#D61F06}{Martin}$ was a greedy man, when he was transporting the apples, he ate one apple for one decameter (quiet an eating machine). But being equally shrewd, he in that way carried the maximum number of apples as he could.

When he reached $\color{limegreen}{Mosaques}$ , he gave the remaining apples to $\color{#20A900}{Marloutte}$ . $\color{#20A900}{Marloutte}$ knew that $\color{#D61F06}{Martin}$ had eaten some apples as it was obvious. On asking, $\color{#D61F06}{Martin}$ told him that for every decametre he travelled he used to eat an apple as he was so greedy that on seeing the apples he could not control himself. He told $\color{#20A900}{Marloutte}$ that the total number of the apples he carried from Mersaills was $5q$ , and he now is left with only $890$ apples.

$\color{#20A900}{Marloutte}$ soon figured out how much apples he began with such that the maximum number of apples he is left with is $890$ . Can you figure out how many apples $\color{#D61F06}{Martin}$ began with such that if he eats one apple for every decameter he travels then the maximum number of apples he could transfer is $890$ ?

ASSUMPTIONS : $\color{#D61F06}{Martin}$ could never control his greed means for every one kilometre he has to and definitely has to eat one apple. Based on his this habit he devised a way to transfer the maximum number of apples he could.

The horse could travel as much as you want it to.

Well, you may as well assume the minimum distance the horse can travel is one decameter in one go. This is for the sake of clarity so that the question reamins logical , realistic and does not go into abstract maths or calculus.

And finally assume everything is in integers, of course positive , that he ate a integer number of bananas each time etc. $\color{#E81990}{REMEMBER}$ It is a big hint to the question and may be the key to it.

And of course this problem is a converse of another problem that I liked which shall be revealed in the solution.

The answer is 47250.

1 solution

Utkarsh Dwivedi
Sep 5, 2014

First have a look at this problem . Clearly to transfer the maximum number of apples you have to make the horse carry maximum number of apples with the consumption of fewest number of apples. The maximum number of apples that the horse can carry is $q$ so what we have to basically do is try to make the horse carry as many apples as possible most of the time. This could be done as first when we are starting we take $q$ apples to one decameter consuming one apple , then we go back one decameter and bring $q$ apples to the place where we had left the other apples and then go back , bring $q$ apples , go back......till we transported all the apples to a distance of one decametre and then we do this again. And as there are $5q$ apples so we need to make $9$ trips to bring all the $5q$ apples to one decameter with a loss of $9$ apples as we had ate them. This goes on till our $q$ apples are consumed leaving us with $4q$ apples. Till then we would have covered $q/9$ kilometers . Then we are left with $4q$ apples so what we do is that we need to now make $7$ trips only to transfer remaining apples to one decameter. We would soon finish with $q$ apples in $q/7$ decametres and so this will continue till we have only $q$ apples left . In that case we shall have no need to go back so we can now continue with our rest of the trip. Now for the maths part . Let us travel $p$ distance till which we had consumed $4q$ apples and left with $q$ apples , then it is evident that $p=q/9 +q/7 +q/5+q/3$ Also after being left with $q$ apples we need to travel $16000 -p$ distance more . So apples remaining after that journey are $q-(9000-p)$ And we know that the left apple are $890$ , so, $q-16000+p=890$ $q-16000+\frac{q}{9}+\frac{q}{7}+\frac{q}{5}+\frac{q}{3}=890$ $q+\frac{744q}{945}=16890$ $q=9450$ So the horse could carry $9450$ apples at a time so we have total apples to be $5\times 9450=47250$

I remember the question was changed to keep the answer. I answered correctly to the original question, but asker made changes to the question instead to fit his originally wrong answer. When I read his posted answer then, it was full of mistakes here and there, not just typos, but the concept, too. This isn't an original question, the asker took one of a common variation, "dramatised" it with M themed names of characters and places and tweaked some numbers, that's all. I found many problems on brilliant adjusted this way and I think it's unfair to earliest solvers. Their correct answers are 'victimized' when people don't admit to the changes they made AFTERWARDS and give some credits back to those who deserves them.

Saya Suka - 1 year, 2 months ago

Martin Management

The answer is 47250.

1 solution

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