My local charity shop was selling off books at 3 paperbacks for $1, 2 hard back for $1 , and $2 for coffee table books, I spent $30 for 30 books and I bought at least one of each book . How many books of each type did I buy?

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- Oct 1st 2009, 12:33 PMciccioAlgebra - Number Theory
My local charity shop was selling off books at 3 paperbacks for $1, 2 hard back for $1 , and $2 for coffee table books, I spent $30 for 30 books and I bought at least one of each book . How many books of each type did I buy?

- Oct 2nd 2009, 01:51 AMCaptainBlack
If $\displaystyle n_p>0$, $\displaystyle n_h>0$ and $\displaystyle n_c>0$ denote the number of paper backs, hard backs and coffee table book bought we have:

$\displaystyle n_p+n_h+n_c=30$

and

$\displaystyle \frac{n_p}{3}+\frac{n_c}{2}+2n_c=30$

Also $\displaystyle n_p$ is divisible by $\displaystyle 3$ and $\displaystyle n_c$ is divisible by $\displaystyle 2$.

Now a bit of trial and error should be sufficient to find the solution.

CB