Difference between revisions of "2001 AMC 12 Problems/Problem 2"
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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> | <math>\text{(A)}\ 2\qquad \text{(B)}\ 3\qquad \text{(C)}\ 6\qquad \text{(D)}\ 8\qquad \text{(E)}\ 9</math> | ||
| − | == Solution == | + | == Solution 1== |
| − | 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 | + | 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 == | ||
{{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}} | {{AMC10 box|year=2001|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:31, 27 October 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 1
Denote 
and 
as the tens and units digit of , respectively. Then 
. It follows that 
, which implies that 
. Since 
, 
. So the units digit of is 
.
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. 
