jsche229 wrote:The question reads as follows.
Answer each question ODD, EVEN, or CANNOT BE DETERMINED. Try to explain each answer using the rules you have just learned in the section. All variables are assumed to be integers unless otherwise indicated.
12. If t=s-3, what is s+t?
I see.
The question is asking us to determine whether or not the given info (t = s - 3) is enough to determine whether s+t is always even, always odd or cannot be determined.
Let's look at 3 different ways to handle this question.
#1
One option is to find some values for t and s that satisfy the condition that t = s - 3, and check whether s + t is odd or even. Let's try a few.
case a) s = 10, t = 7, in which case s + t = 17 (ODD)
case b) s = 4, t = 1, in which case s + t = 5 (ODD)
case c) s = -5, t = -8, in which case s + t = -13 (ODD)
case d) s = 7, t = 4, in which case s + t = 11 (ODD)
.
.
.
It
looks like s + t will always be odd. We could try some more possible values, and each time we see that s + t is odd, we can be more certain of our conclusion.
Having said that, listing possible cases doesn't provide conclusive evidence.
#2
For conclusive evidence, let's work with the given info (t = s - 3)
Take t = s - 3
Subtract s from both sided to get: t - s = -3
Since -3 is odd, we know that the
difference between t and s is odd.
From this, we can conclude that the
sum of t and s will also be odd.
This is a rule.
#3
Here's one last approach.
t = s - 3
In other words, t = s - (some odd number)
Let's consider two possible cases (s is odd, and s is even)
case a: s is even: we get t = (even) - (some odd number) = odd
So, s is even, and t is odd, which means
s + t = odd
case b: s is odd: we get t = (odd) - (some odd number) = even
So, s is odd, and t is even, which means
s + t = odd
In both case,
s + t = odd, so it must be the case that
s + t is always odd
Cheers,
Brent
