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2001 AMC 10 Problems/Problem 6: Difference between revisions

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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>?
two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit of <math>N</math>?


<math>\text{(A)}\ 2\qquad \text{(B)}\ 3\qquad \text{(C)}\ 6\qquad \text{(D)}\ 8\qquad \text{(E)}\ 9</math>
<math>\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 9</math>


== 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>.
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>.
 
==Video Solution by Daily Dose of Math==
 
https://youtu.be/t6AxKdlda7I?si=ZIZP5ZyA0JN-Bgf9
 
~Thesmartgreekmathdude


== See Also ==
== See Also ==
{{AMC10 box|year=2001|num-b=5|num-a=7}}
{{AMC12 box|year=2001|num-b=1|num-a=3}}
{{AMC12 box|year=2001|num-b=1|num-a=3}}
{{AMC10 box|year=2001|num-b=5|num-a=7}}
{{MAA Notice}}
{{MAA Notice}}
[[Category: Introductory Number Theory Problems]]

Latest revision as of 16:13, 18 October 2025

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

Video Solution by Daily Dose of Math

https://youtu.be/t6AxKdlda7I?si=ZIZP5ZyA0JN-Bgf9

~Thesmartgreekmathdude

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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