I would approach the problem as :

rs + r
=r(s+1)

In the problem it is mentioned that rs + r is odd. Now by number system properties we know that :

odd * odd = odd (i.e. 3*9=27)

From this we can deduce that r and s+1 both are odd numbers. So the answer would be - (b)

Now s and s+1 are consecutive integers. This makes s an even number as s+1 is an odd number.

Hope this helps...

rog wrote:I had trouble with this question on my last practice exam. Any help on how to systematically approach this kind of question would be greatly appreciated.

If r and s are integers and rs + r is odd, which of the following must be even?

a) r
b) s
c) r + s
d) rs - r
e) r^2 + s

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If r(s+1) is odd that means r and s+1 have to be odd, thereby meaning that s has to be even. (even+1 would make an odd). So r = odd and s = even.
1. Odd. Incorrect.
2. even. Correct.
3. Odd+Even = Odd so incorrect
4. r(s-1) = odd(even-1) = odd*odd = odd , so incorrect.
5. r^2+s=odd^2+even=odd+even=odd, so incorrect.

B(2) is the answer

Cheers
Ami/-

rog wrote:I had trouble with this question on my last practice exam. Any help on how to systematically approach this kind of question would be greatly appreciated.

If r and s are integers and rs + r is odd, which of the following must be even?

a) r
b) s
c) r + s
d) rs - r
e) r^2 + s

Plug in values for r and s that satisfy the condition that rs+r is odd.
Plug these values for r and s into the answer choices.
Eliminate any answer choice that isn't even.

Let r=3 and s=2 so that rs+r = (3*2)+3 = 9, which is odd.

A: r=3. Not even. Eliminate A.
B: s=2. Even. Hold onto B.
C: r+s = 3+2 = 5. Not even. Eliminate C.
D: rs-r = (3*2)-3 = 3. Not even. Eliminate D.
E: r² + s = 3²+2 = 11. Not even. Eliminate E.

Only answer choice B remains.
The correct answer is B.

If more than one answer remained, we would:

Plug in different values for r and s that satisfy the condition that rs+r is odd.
Plug these new values for r and s into the remaining answer choices.
Eliminate any answer choice that isn't even.
Continue this process until only one answer choice is left, being sure to try different combinations of even and odd values.
.