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A pen and a pencil cost an integer number of cents. It is known that 175 pencils cost more than 125 pens but less than 126 pens. Prove that 3 pencils and a pen cost more than $1.

Let a pencil cost p cents and a pen cost q cents. $$126q>175p>125q$$ $$126q\geq 175p+1 \\ 175p\geq125q+1$$ Then, $$126q\geq 125q+2$$ $$q\geq2$$ So $$175p\leq 126q-1\\ 175p \geq 125q+1$$ Combined with the last inequality $$251\leq 175p\leq 251$$ $$p=\frac{251}{175}$$ But that means $$3p+q\geq \frac{1103}{175} <<< 100$$ How can I get stonger bounds on q, or did I do something wrong?

Robert Soupe
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Baker013273213
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As you say, $$125q<175p<126q\\\frac{125}{175}<\frac pq<\frac{126}{175}$$ Also, both $p$ and $q$ are positive integers, so the problem has been reduced to finding the rational number $\frac pq$ with smallest denominator such that $\frac pq\in\left(\frac {125}{175},\frac{126}{175}\right)$.

Personally, I found this with a quick computer program. But checking Minimal $ab$ for Rational Number $a/b$ in an Interval, the accepted answer contains a solid algorithm for calculating the solution essentially based on the continued fractions of the two endpoints. In either case, $(p,q)=(23,32)$ is the minimal solution.

Even in this smallest case, $$3p+q=3(23)+32=101$$ so the total cost must be larger than one dollar.