Difference between revisions of "1999 AMC 8 Problems/Problem 1"
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==Solution== | ==Solution== | ||
− | + | Simplifying the given expression, we get: <math>(6?3)=2</math> | |
− | <math> (6?3) | ||
− | + | At this point, it becomes clear that it should be <math> \div,</math> so the answer is <math> \boxed{A} </math>. | |
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− | At this point, it becomes clear that it should be <math> \div, \boxed{A} </math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=1999|before=First<br />Question|num-a=2}} | {{AMC8 box|year=1999|before=First<br />Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:23, 16 July 2018
Problem
To make this statement true, the question mark between the 6 and the 3 should be replaced by
Solution
Simplifying the given expression, we get:
At this point, it becomes clear that it should be so the answer is .
See Also
1999 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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.