Difference between revisions of "2009 AMC 10B Problems/Problem 7"
Coltsfan10 (talk | contribs) (→Solution) |
Whitelisted (talk | contribs) m (→Problem) |
||
Line 1: | Line 1: | ||
{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #7]] and [[2009 AMC 12B Problems|2009 AMC 12B #6]]}} | {{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #7]] and [[2009 AMC 12B Problems|2009 AMC 12B #6]]}} | ||
− | + | why do u ned de anser | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
== Solution == | == Solution == |
Revision as of 09:19, 24 May 2021
- The following problem is from both the 2009 AMC 10B #7 and 2009 AMC 12B #6, so both problems redirect to this page.
why do u ned de anser
Solution
The three operations can be performed on any of orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions are in fact all distinct. So the answer is , which is choice .
See also
2009 AMC 10B (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 10 Problems and Solutions |
2009 AMC 12B (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.