Difference between revisions of "2020 AMC 8 Problems/Problem 7"
Itsmemasters (talk | contribs) (→See also) |
|||
| Line 6: | Line 6: | ||
{{AMC8 box|year=2020|num-b=6|num-a=8}} | {{AMC8 box|year=2020|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | |||
| + | The thousands place (first digit) has to be a 2 (2020-2400). | ||
| + | Since the thousands digit is 2, the next digit must be a 3 (not 4 or onwards because that will go over the range given). | ||
| + | |||
| + | The next digit has to be from 4, 5, 6, 7, or 8. For each of the cases, you get a total of 15 possibilities, which gives you the answer C. | ||
| + | |||
| + | |||
| + | ~itsmemasterS | ||
Revision as of 02:52, 18 November 2020
Problem 7
How many integers between 
and 
have four distinct digits arranged in increasing order? (For example, 
is one integer.)
See also
| 2020 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
The thousands place (first digit) has to be a 2 (2020-2400).
Since the thousands digit is 2, the next digit must be a 3 (not 4 or onwards because that will go over the range given).
The next digit has to be from 4, 5, 6, 7, or 8. For each of the cases, you get a total of 15 possibilities, which gives you the answer C.
~itsmemasterS
