<p>I got this question from the official 08-09 SAT Practice Test. Section 2 #15.</p>
<li>If p and n are integers such that p > n > 0 and p^2-n^2=12, which of the following can be the value of p-n?</li>
</ol>
<p>I. 1
II. 2
III. 4</p>
<p>a. I only
b. II only
c. I and II only
d. II and III only
e. I, II, and III </p>
<p>Can someone show me a fast way to do this problem? I had to skip this problem and spend about 4 minutes after I finished the rest of the section to find the answer. The answer is b.</p>
<p>you can break p^2-n^2 down to (p-n)(p+n) because it is difference of cubes.</p>
<p>now you begin the guessing and checking...</p>
<p>if (p-n) were to equal 1, then that would mean p+n would have to equal 12, but can you think of two numbers that have a sum of 12 but at the same time a difference of 1? 6 and 7=13 so no...5 and 6 =11 so no.</p>
<p>same thing with choices II and III and you should arrive at the answer B.</p>
<p>I know, I did do that, but it takes a long time. I was wondering if anyone knew a fast 15-30 second answer. Like if there's some relationship that I should see right away but didn't. Thanks for your input though =]</p>
<p>With practice you will notice that p-n and p+n are both odd or both even. For p-n=1, p+n=12 wrong. for p-n=2, p+n=6 right, for p-n=4, p+n=3 wrong. You can also eliminate the last choice because for positive numbers p-n<p+n, but="" 4="">3. It doesn't take more than 15 seconds.</p+n,></p>
<p>^ wow. thats a good one =p
i have to admit i got stuck on this one last time too, basically i had to skip it and come back, and i think it got it wrong lol...
its very keen to spot that both odd/even thing. nice =p</p>
<p>May be this might be helpful.
since p>n>0 both p and n are positive integers.
p has to be greater than 0 and p^2 has to be more than 12
then by plugging in the values we can tell p=4 and n=2
hence, p-n=2</p>