Difference between revisions of "MIE 2016/Problem 2"
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There are <math>k=6</math> solutions and <math>6\leq k\leq8</math>, giving <math>\boxed{D)}</math> | There are <math>k=6</math> solutions and <math>6\leq k\leq8</math>, giving <math>\boxed{D)}</math> | ||
+ | ~Windigo | ||
==See Also== | ==See Also== |
Revision as of 11:08, 10 September 2020
Problem 2
The following system has integer solutions. We can say that:
(a)
(b)
(c)
(d)
(e)
Solution 1
Objective:
We can solve this problem in two steps: First, we solve for the range of , then combine it with the range of
to get a compound inequality which we can use to find all possible integer solutions.
Step 1
We first find the range of the inequality .
We now simplify the inequality:
Case 1:
Factoring, we get
Now,
can be greater than
or less than
.
But in this case,
, and this further restricts our solutions.
So, for the case where
, our solutions are
Case 2:
We have in this case that
, but the case statement further restricts our solutions.
For this case, the solutions are
Step 2
Now, we know the solutions for : in the first case, where
, the integer solutions are
In the second case, where , the only integer solution is
The union of these two cases gives .
There are solutions and
, giving
~Windigo