The sum of the prices of a pen, an eraser and a notebook is $100$ rupees. The price of a notebook is greater than the price of two pens. The price of three pens is greater than the price of four erasers and the price of three erasers is greater than the price of a notebook. If all of the prices are in integers, find each of their prices.
I was encountered with the problem quite a long time ago and today the problem again came to my notice. That time I solved the problem in the following way:
Let the prices of a pen, an eraser and a notebook be $p$, $e$ and $n$ respectively. Then we have the following: $$p+e+n=100\\ n>2p\\ 3p>4e\\ 3e>n$$
We have $p>\frac43e$ and $n>2p>\frac83e$. So, $$\frac43e+e+\frac83e<100\\ \implies e<20$$ Similarly, we get $$p\leq 27 \ \ \ \ \text{and}\\ n\leq 56$$ Setting the values of $e$ and $p$ and with a bit brute forcing, I got $e=19$, $p=26$ and $n=55$ which I think is the only solution.
Is the solution correct? And is there some other way which gives the result directly i.e. without brute forcing? Also as we have inequalities involved in our problem, can we use some known inequalities like AM-GM, Cauchy-Schwarz?
$p+e+n=100\\ n>2p\\ 3p>4e\\ 3e>n$
As you found $n \leq 56, p \leq 27$
Now we write upper bound of $p$ and $e$ in terms of $n$.
$p \lt \frac{n}{2}, e \lt \frac{3n}{8}$
So, $\frac{n}{2} + \frac{3n}{8} + n \gt 100 \implies n \gt \frac{160}{3}$
So, $54 \leq n \leq 56$. Also, $3e \gt n \implies e \geq 19$ and we already have $e \lt 20$. So $e = 19$ is the only solution.
You can also use $p \gt \frac{4e}{3} \implies 26 \leq p \leq 27$.
$p = 27, n = 54$ does not satisfy $n \gt 2p$. So there is only one solution as you wrote.