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

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Please navigate here for the solution: http://www.artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_2
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{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}}
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== Problem ==
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Let <math>P(n)</math> and <math>S(n)</math> denote the product and the sum, respectively, of the digits
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of the integer <math>n</math>. For example, <math>P(23) = 6</math> and <math>S(23) = 5</math>. Suppose <math>N</math> is a
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two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit of <math>N</math>?
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<math>\text{(A)}\ 2\qquad \text{(B)}\ 3\qquad \text{(C)}\ 6\qquad \text{(D)}\ 8\qquad \text{(E)}\ 9</math>
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== Solution ==
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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{(\text{E})9}</math>.
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== See Also ==
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{{AMC12 box|year=2001|num-b=1|num-a=3}}
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{{AMC10 box|year=2001|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 18:28, 28 January 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 $\boxed{(\text{E})9}$.

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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