I am apple @_@

Probability Level 3

There are $4$ types of boxes of $\color{#D61F06}{\mathrm{apples}}$ ,

$\bullet \quad$ boxes containing $40$ $\color{#D61F06}{\mathrm{apples}}$ ,

$\bullet \quad$ boxes containing $16$ $\color{#D61F06}{\mathrm{apples}}$ ,

$\bullet \quad$ boxes containing $56$ $\color{#D61F06}{\mathrm{apples}}$ ,

$\bullet \quad$ boxes containing $32$ $\color{#D61F06}{\mathrm{apples}}$ ,

Then find the maximum number of $\color{#D61F06}{\mathrm{apples}}$ less than $2014$ which can be given by giving some boxes of some of the types stated above.

Details and assumptions :-

You are allowed to give 0 boxes of a certain type(s), and thus, find the maximum number.

This problem is a part of the set Vegetable Combinatorics , try other problems too !

The answer is 2008.

1 solution

Aditya Raut
Aug 8, 2014

See that $gcd(40,16,56,32)=8$ , hence the number of apples will always be a multiple of $8$ .

The maximum multiple of $8$ less than $2014$ is $2008$ , so only thing remains is to see if we can give $2008$ apples using the conditions.

This is same as finding if we can give $\dfrac{2008}{8}= 251$ apples by using boxes of $5,2,7,4$

Thus is true, if you take $48$ boxes of $5$ and a box of $4$ with one of $7$ , you are getting it! This holds for many more combinations, but we have proved that it holds . Thus the maximum number less than 2014 which can be given by boxes is $\boxed{2008}$

i thought that the answer was remainder and kept typing my answer as 6 ! so silly

Apurv Rajput - 6 years, 10 months ago

I am apple @_@

The answer is 2008.

1 solution

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