Difference between revisions of "2001 AMC 10 Problems/Problem 6"
(→Problem) |
|||
(2 intermediate revisions by 2 users not shown) | |||
Line 9: | Line 9: | ||
== 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{(\ | + | 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}} | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:12, 15 July 2024
- 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 and
denote the product and the sum, respectively, of the digits
of the integer
. For example,
and
. Suppose
is a
two-digit number such that
. What is the units digit of
?
Solution
Denote and
as the tens and units digit of
, respectively. Then
. It follows that
, which implies that
. Since
,
. So the units digit of
is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.