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Difference between revisions of "2001 AMC 10 Problems/Problem 6"

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<math> \textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9 </math>
 
<math> \textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9 </math>
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== Solution ==
 
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Let <math> N=10a+b </math>. We want to find a number such that <math> ab+a+b=10a+b </math>.
 
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If we subtract the quantity <math> a+b </math> from both sides, we are left with
 
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<math> ab=9a </math>.
 
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We can also divide both sides by a and we are left with
 
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the units digits <math> b </math> is <math> \boxed{\textbf{(E) }9} </math>.
 

Revision as of 15:06, 16 March 2011

Problem

Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$. For example, $P(23) = 6$ and $S(23) = 5$. Suppose $N$ is a two-digit number such that $N = P(N) + S(N)$. What is the units digit of $N$?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

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