SOLUTION: Charlie has a collection of books that he wishes to display in a narrow bookcase with shelves of width 56 cm. The thickest books are no more than 16 cm wide and, when placed side

Algebra ->  Length-and-distance -> SOLUTION: Charlie has a collection of books that he wishes to display in a narrow bookcase with shelves of width 56 cm. The thickest books are no more than 16 cm wide and, when placed side      Log On


   

Question 1166355: Charlie has a collection of books that he wishes to display in a narrow
bookcase with shelves of width 56 cm. The thickest books are no more
than 16 cm wide and, when placed side by side, the entire collection takes
up 2.4m. Find, with justification, the minimum number of shelves required
to guarantee that all of the books can be displayed in the bookcase.
Found 3 solutions by solver91311, ikleyn, math_helper:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The maximum number of shelves that could be required is 5: 240 cm divided by 48 cm (the width of three 16 cm wide books, the most that could fit on one shelf) given that all of the books measured 16 cm.

But even in the optimum case where the assortment of thicknesses was such that it was possible to fill four shelves completely, 240 divided by 56 is 4 with a remainder, so you would have to put 16 centimeters worth of books on the fifth shelf, so the minimum number is also 5.

John

My calculator said it, I believe it, that settles it


I > Ø

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.


            The solution by tutor  @solver91311  is not correct.
            To demonstrate it,  I will construct a contradictory example. 


Let all the books are 15 cm wide.


Then the total number of books is  240%2F15 = 16,

and we can place  ONLY  3  such books at each shelf.



So, having 5 shelves, we can place only  3*5 = 15 such books, and we need then

the 6-th shelf for the 16-th book.


            Now, after completing this counter-example,  I can solve the problem in full.
            My statement is that  6  shelves is always enough. 


1)  In the 1-st shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



2)  In the 2-nd shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



3)  In the 3-rd shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



4)  In the 4-th shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



5)  In the 5-th shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



6)  In the 6-th shelf I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



So, I can fill at least 40 cm of 56 cm in each of 6 shelves.



Taken together,  6 times 40 cm comprise  2 m 40 cm,

which means that ALL the books will be placed in 6 shelves.

        ***********************************
            THE PROOF IS COMPLETED.
        ***********************************


It is a  TRUE  Math  Olympiad level  Math  problem  (!)



Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!

Thanks tutor ikleyn for that explanation.