<p>If p and n integers such that p>n>0 and p^2 - n^2 = 12 which fo the following can be the value of p-n?
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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>4 and 2 is all i can think of</p>
<p>(4^2)-(2^2)=12</p>
<p>4-2=2</p>
<p>B</p>
<p>(c)</p>
<p>answer explained:
p^2-n^2=12
we see that there is a number squared minus a number squared
thus simplified as:
(p-n)(p+n)=12</p>
<p>the question is asking which of the folowing can be p-n</p>
<p>we need to check by plugging #'s:
if I is true then :
p-n=1
p+n=12
(we solve the set of two equations)
2p=13 :2
p=6.5
6.5-n=1
n=5.5
6.5>5.5>0 I is good</p>
<p>if II is true then :
p-n=2
p+n=6
(we solve the set of two equations)
2p=8 :2
p=4
4-n=2
n=2</p>
<p>4>2>0 II is good</p>
<p>if III is true then :
p-n=4
p+n=3
(we solve the set of two equations)
2p=7 :2
p=3.5
3.5-n=4
n=-0.5</p>
<p>3.5>0>-0.5 III isn’t good </p>
<p>(C) is the right answer</p>
<p>P and N must be integers ^</p>
<p>and P>N>0</p>
<p>p²-n² = 12
(p-n)(p+n) = 12</p>
<p>Factorization of 12 : 1, 2, 3, 4, 6, 12
pairs: 1x12, 2x6, 3x4
You can try them out and find out that only 2 is possible.
Answer is B</p>
<p>If you’re good at logic, you can use this as well since it is given that p and n are both integers:
even + or - even = even;
even + or - odd = odd;
odd + or - odd = even.</p>
<p>oh yea I missed that. (b) then. Thanks.</p>
<p>All in all this is the way he should face this problem. once mastered, it shouldn’t take more the 20 secs.</p>
<p>I agree with Kieran0696 that the answer is B. Here’s why:</p>
<p>p^2 - n^2 = 12 is the same as (p-n)(p+n)=12
We know that (p-n) must be at least one of the three values, so we match them up:</p>
<p>
</p>
<p>With that, you end up with “B. II only”</p>
<p>
I actually think this problem takes a bit more time than others.
Without using the logic of addition and subtraction of even and odds, it seems like it will take at least 1 minute to solve this. The work that is required is very tedious.</p>
<p>I do agree with you on that the way you approached the problem is probably the most common and the easiest one.</p>
<p>Another tip for this problem is understanding that since p>n>0, (p+n) is always greater than (p-n).</p>
<p>(p + n)(p - n) = 12;
(p + n) * 1 = 12;</p>
<p>(p + n) = 12;</p>
<p>What number minus each other = 1 and plus each other = 12?
It must be then a number that has following format</p>
<p>p + p-1 = 12; p - p - 1 = 12;</p>
<p>Ofcourse no number has following format;</p>
<p>(p + n)(p - n) = 12;
(p + n)(2) = 12;
(p + n) = 6;</p>
<p>p + n = 6;
+
p - n = 2; </p>
<p>2p = 8;
p = 6;</p>
<p>6 - n = 2;</p>
<p>-n = -4;
n = 4;</p>
<p>Lets try now the third number:-</p>
<p>p^2 - n^2 = 12; </p>
<p>(p + n)(p - n) = 12;
(p + n)(4) = 12;
(p + n) = 3;</p>
<p>p + n = 3;
+
p - n = 4;</p>
<p>2p = 7;
p = 3.5,since they can only be integers so the answer is II only;</p>
<p>I hope that helped by the way math just improves as you solve more questions,but you have to understand the logic behind the way how its solved,because you might use that logic to solve another question :).</p>
<p>@JefferyJung</p>
<p>Nice logic JefferyJung its much faster than using system of equation to solve for the value
factorization would be the fastest way to solve this would save some time.</p>
<p>
</p>
<p>I think this problem would take approximately 1 minute using system of equations. However, with factorization + even, odd method, it is certainly possible to solve this problem within 30 seconds.</p>
<p>The biggest struggle which many students face with more difficult SAT math problems is understanding how to approach the problem. If you begin one step for those students, they fully understand what to do next. Finding the appropriate and the most effective method of solving requires sound basics…</p>
<p>@JefferyJung</p>
<p>Well yeah in order to have that kind of reaction in exam to like pick the best method one must solve many questions in math in order to first have many tools in his toolbar to pick the best logic to use in order to solve a problem.</p>