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

Revision as of 08:19, 20 August 2019

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 $N$ ?

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

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 $\boxed{(\text{E})\ 9}$ .

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

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

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

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 == See Also ==
+{{AMC10 box|year=2001|num-b=5|num-a=7}}
 {{AMC12 box|year=2001|num-b=1|num-a=3}}
-{{AMC10 box|year=2001|num-b=5|num-a=7}}
 {{MAA Notice}}

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

Revision as of 08:19, 20 August 2019

Problem

Solution

See Also