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Problem 104

AMC 10/12, early questions

Number theory Difficulty 3.5 Find the answer

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

Your answer

Official solution

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 $\boxed{\textbf{(C) }10.5}.$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.