<p>For all numbers x and y, let the operation (square) be defined by x (square) y= xy-y. If a and b are positive integers, which of the following can be equal to zero?</p>
<p>I. a (square) b
II. (a+b) (square) b
III. a (square) (a+b)</p>
<p>A. I only
B. II only
C. III only
D. I and II
E. I and III</p>
<p>Correct answer is E, can someone explain this problem fully. I really have trouble with these when they say let this___ be defined by the operation stuff. Thanks to anyone who helps!</p>
<p>The first step is to use the “defined” operation and replace the terms. </p>
<p>This means to apply the operation x (square) y = xy-y. </p>
<p>I. a (square) b ===> ab - b. This becomes b(a-1)</p>
<p>II. (a+b) (square) b ===> (a+b)b - b. This becomes b(a+b-1)</p>
<p>III. a (square) (a+b) ===> a(a+b) - (a+b). This becomes (a+b) (a-1)</p>
<p>Since a and b have to positive, they cannot be zero. In the above terms, we have isolated (a-1) … this means that if a = 1, there is a solution that can be 0, namely b(1-1) = b(0) or 0. This applies to both I and III. </p>
<p>For II, a+b is always greater than 1. This means that (a+b-1) cannot be zero, and neither can II = 0. </p>
<p>Well, simplifying it, you get
a^3<em>b^2=432
or
a</em>(ab)^2=432
Factor 432, and you get
2<em>2</em>2<em>2</em>3<em>3</em>3
The possible factors that are perfect squares are 4, 9, 16, 36, and 144 One of these is (ab)^2</p>
<p>Respectively, this means that a is 108, 48, 27, 12, and 3. A must be less than ab(whose possible values are 2, 3, 4, 6, and 12). Thus, a must be 3, which means (ab)^2 is 144, which means ab is 12.</p>
<p>thanks, i still think i would never be able to figure that out on the test in real time. Baffles me how people get 2400’s. That problem alone takes like 3 minutes right? Lol, im just dumb, i just dont get how you would know when to simplify, for ex. when you have a^3 b^2= 432. How do you know to simplify to a *(ab)^2? It just comes to you? I think at that point I would just start plugging in numbers…lol</p>