Competition problem — Algebra — Olympiad Maths Prep

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Algebra Difficulty 5.7 AIME, harder Prove it Ukraine

It is given that the bookshelf can fit $9$ of the same thick books, but the $10$ th one will not fit anymore. Similarly, it can hold $15$ of the same thin books, but $16$ th will not fit anymore. Is it possible for that shelf to hold simultaneously:

a) $6$ thick and $5$ thin books?

b) $7$ thick and $5$ thin books?

Solution

Let us re-write the statement of the problem as follows. Let us denote the length of the shelf by $S$ , the width of the thick book by $x$ and the width of the thin book by $y$ . Then, we have the conditions:
$9x \leq S < 10x \text{ and } 15y \leq S < 16y \Leftrightarrow \frac{1}{10}S < x \leq \frac{1}{9}S \text{ and } \frac{1}{16}S < y \leq \frac{1}{15}S.$

a) Let's see what is the maximum width of the space these books would take:
$6x + 5y \leq \frac{6}{9}S + \frac{5}{15}S = S.$

b) Analogously, we show that this set of books cannot fit on the shelf:
$7x + 5y > \frac{7}{10}S + \frac{5}{16}S = \frac{56+25}{80}S > S.$

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