1972 AHSME Problems/Problem 33

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Problem 33

The minimum value of the quotient of a (base ten) number of three different non-zero digits divided by the sum of its digits is

$\textbf{(A) }9.7\qquad \textbf{(B) }10.1\qquad \textbf{(C) }10.5\qquad \textbf{(D) }10.9\qquad  \textbf{(E) }20.5$


The answer we are looking for can be expressed as $\dfrac{100a+10b+c}{a+b+c}$. This is equivalent to $1 + \dfrac{99a+9b}{a+b+c}$. Because we are trying to minimize our solution, we set $c$ = $9$, so we have $1 + \dfrac{99a+9b}{a+b+9}$. This is equal to $1 + \dfrac{9a+9b+81}{a+b+9} + \dfrac{90a-81}{a+b+9}$, which simplifies to $10+ \dfrac{90a-81}{a+b+9}$. Since each digit is unique, we set $b$ to $8$, leaving us with $10 + \dfrac{90a-81}{a+17}$. Clearly, $a$ should be minimized, so $a = 1$ and our answer is \[\boxed{\textbf{(C) }10.5}.\]