Difference between revisions of "2011 AMC 12A Problems/Problem 20"
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− | And since <math>15000 < 19295 < 20000 \to 5000(3) < 19295 < 5000(4)</math>, we find that <math>k = 3</math>, which is \boxed{(\textbf{C})}. | + | And since <math>15000 < 19295 < 20000 \to 5000(3) < 19295 < 5000(4)</math>, we find that <math>k = 3</math>, which is <math>\boxed{(\textbf{C})}</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}} | {{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}} |
Revision as of 13:08, 10 February 2011
Problem
Let , where
,
, and
are integers. Suppose that
,
,
,
for some integer
. What is
?
Solution
From , we know that
. From the first inequality:
Since must be an integer, it follows that
. Similarly, from the second inequality:
And it follows that . We now have a system of three equations. Solving it gives us
. From this, we find that
And since , we find that
, which is
.
See also
2011 AMC 12A (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 12 Problems and Solutions |