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

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== Solution ==
 
== Solution ==
Denote <math>a</math> and <math>b</math> as the tens and units digit of <math>N</math>, respectively. Then <math>N = 10a+b</math>. It follows that <math>10a+b=ab+a+b</math>, which implies that <math>9a=ab</math>. Since <math>a\neq0</math>, <math>b=9</math>. So the units digit of <math>N</math> is <math>\boxed{(\text{E})9}</math>.
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Denote <math>a</math> and <math>b</math> as the tens and units digit of <math>N</math>, respectively. Then <math>N = 10a+b</math>. It follows that <math>10a+b=ab+a+b</math>, which implies that <math>9a=ab</math>. Since <math>a\neq0</math>, <math>b=9</math>. So the units digit of <math>N</math> is <math>\boxed{(\textbf{E})\ 9}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 22:12, 27 July 2016

The following problem is from both the 2001 AMC 12 #2 and 2001 AMC 10 #6, so both problems redirect to this page.

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 $N$?

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

Solution

Denote $a$ and $b$ as the tens and units digit of $N$, respectively. Then $N = 10a+b$. It follows that $10a+b=ab+a+b$, which implies that $9a=ab$. Since $a\neq0$, $b=9$. So the units digit of $N$ is $\boxed{(\textbf{E})\ 9}$.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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