Difference between revisions of "2002 AMC 12P Problems/Problem 7"
(→Solution) |
(→See also) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| + | {{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #7]] and [[2002 AMC 10P Problems|2002 AMC 10P #20]]}} | ||
| + | |||
== Problem == | == Problem == | ||
How many three-digit numbers have at least one <math>2</math> and at least one <math>3</math>? | How many three-digit numbers have at least one <math>2</math> and at least one <math>3</math>? | ||
| Line 26: | Line 28: | ||
This case is almost identical to the second case, just swap the <math>2</math>s with <math>3</math>s and <math>3</math>s with <math>2</math>s in the reasoning and its the same, giving us an additional <math>19</math> cases. | This case is almost identical to the second case, just swap the <math>2</math>s with <math>3</math>s and <math>3</math>s with <math>2</math>s in the reasoning and its the same, giving us an additional <math>19</math> cases. | ||
| − | + | Adding up all of these cases gives <math>14+19+19=52</math> cases, or <math>\boxed{\textbf{(A) }52}.</math> | |
== See also == | == See also == | ||
| + | {{AMC10 box|year=2002|ab=P|num-b=19|num-a=21}} | ||
{{AMC12 box|year=2002|ab=P|num-b=6|num-a=8}} | {{AMC12 box|year=2002|ab=P|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:20, 14 July 2024
- The following problem is from both the 2002 AMC 12P #7 and 2002 AMC 10P #20, so both problems redirect to this page.
Problem
How many three-digit numbers have at least one 
and at least one 
?
Solution
We can do this problem with some simple case work.
Case 1: The hundreds place is not or 
This means that the tens place and ones place must be and respectively or and respectively. This case covers 
and 
so it gives us 
cases.
Case 2: The hundreds place is 
This means that must be in the tens place or ones place. Starting with cases in which the tens place is not , we get 
and 
With cases in which the tens place is , we have 
, or 
more cases. This gives us 
cases.
Case 3: The hundreds place is
This case is almost identical to the second case, just swap the s with s and s with s in the reasoning and its the same, giving us an additional 
cases.
Adding up all of these cases gives 
cases, or 
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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. 
