Difference between revisions of "2013 AMC 12A Problems/Problem 25"
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When <math>a\leq 0</math>, there is no restriction on <math>b</math> so there are <math>11\cdot 21 = 231</math> pairs; | When <math>a\leq 0</math>, there is no restriction on <math>b</math> so there are <math>11\cdot 21 = 231</math> pairs; | ||
− | when <math>a > 0</math>, there are <math>2( | + | when <math>a > 0</math>, there are <math>2(1+4+9+10+10+10+10+10+10+10)=2(84)=168</math> pairs. |
So there are <math>231+168=399</math> in total. | So there are <math>231+168=399</math> in total. |
Revision as of 14:28, 28 January 2014
Problem
Let be defined by
. How many complex numbers
are there such that
and both the real and the imaginary parts of
are integers with absolute value at most
?
Solution
Suppose . We look for
with
such that
are integers where
.
First, use the quadratic formula:
Generally, consider the imaginary part of a radical of a complex number: , where
.
.
Now let , then
,
,
.
Note that if and only if
. The latter is true only when we take the positive sign, and that
,
or ,
, or
.
In other words, for all ,
satisfies
, with a unique solution
. Therefore we need to count the number of ordered pairs
such that integers
, and that
.
When , there is no restriction on
so there are
pairs;
when , there are
pairs.
So there are in total.
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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 |
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