<p>Ok, so here's how the question goes :</p>
<p>If p and n are integers such that p > n > 0 and </p>
<p>p2(as in p square) - n2 = 12, which of the following can be the value of p-n :</p>
<p>I. 1
II. 2
III. 4</p>
<p>(a) I only</p>
<p>(b) II only</p>
<p>(c) I and II only</p>
<p>(d) II and III only</p>
<p>(e) I, II and III</p>
<p>Plz explain the answer!! Thankss a ton! :)</p>
<p>Well since your answers are all integers, you know that your numbers p and n are gonna be perfect squares.</p>
<p>So start thinking in your head of the perfects squares and ‘p’ has to be greater than ‘n’ and they’re both positive.</p>
<p>2 squared - 1 sqaured = 2 NO!
3 squared is 9 so thats not high enough
4 squared is 16 … minus 2 squared = 12!</p>
<p>if you think of the rest of your perfect squares nothings going to be able to subtracted by another perfect square to give you 12.</p>
<p>So I would say the answer is (4-2=2) 2 only</p>
<p>hmmmm
wouldn’t it be like p^2-n^2=12</p>
<p>so (p+n)(p-n)=12</p>
<p>so for 2, 4 and 2 where (4+2)(4-2)= 12 so 2 is right </p>
<p>so yes answers only work for 2</p>
<p>Here is a complete algebraic solution:</p>
<p>p^2-n^2 = (p+n)(p-n) = 12</p>
<p>If p-n = 1, then p+n = 12. Adding these equtions gives:</p>
<p>p-n=1
p+n=12
2p = 13. So p is NOT an integer. </p>
<p>If p-n=2, then p+n=6. Adding these equations gives 2p=8 and p = 4. So n = 2 . So this works.</p>
<p>If p-n=4, then p+n=3. So 2p = 7, and p is not an integer.</p>
<p>Thus the answer is II only,choice (B).</p>
<p>On the SAT I would probably answer the question by starting with n=1, and solving for p.</p>
<p>n=1 p^2 - 1=12. So p^2=13, no good.
n=2 p^2-4=12. So p^2=16. GOOD! p=4, and p-n = 2.
n=3 p^2-9=12. So p^2=21, no good.
n=4 p^2-16=12. So p^2=28, no good.</p>
<p>Check a few more, and then take an educated guess that the answer is (B).</p>
<p>Using Table on TI-8X.</p>
<p>Answer options I, II, and III translate:
I. p=n+1
II. p=n+2
III. p=n+4</p>
<p>Then p^2 - n^2 would be respectively equal to
(n+1)^2 - n^2
(n+2)^2 - n^2
(n+4)^2 - n^2</p>
<p>Let’s find when p^2 - n^2 = 12.</p>
<p>Set the Table (TBLSET) for positive integers x.</p>
<p>I. Enter in Y-editor
Y1=(x+1)^2-x^2 and check the Table.</p>
<p>II. Change Y1=(x+1)^2-x^2 to
Y1=(x+2)^2-x^2 and check the Table.</p>
<p>III. Change Y1=(x+2)^2-x^2 to
Y1=(x+4)^2-x^2 and check the Table.</p>
<p>Y1=12 only for
Y1=(x+2)^2-x^2 (at x=2).
Answer (B).</p>
<p>Yet another - more efficient albeit less obvious - approach.</p>
<p>First, simple properties:
even ± even = even;
even ± odd = odd;
odd ± odd = even.</p>
<p>p^2-n^2=12
(p-n)(p+n)=12
12=1x12=2x6=3x4
Since (p-n) and (p+n) have the same parity (both are either even or odd), only
2x6=12 is possible, thus p-n=2.</p>
<p>Can someone PLEASE help me with this question. Its such an easy question i dunno why i cant get the right answer.</p>
<p>a, 3a, …</p>
<ol>
<li>The first term in the sequence above is a, and each term after is 3 times the preceding term. If the sum of the first 5 terms is 605, what is the value of a?</li>
</ol>
<p>What i did:</p>
<p>∑first5terms = a + 3a + 9a + 27a + 81a = 121a = 605 a = 5</p>
<p>But it says the answer is 6??? How?</p>
<p>This is from p597 of Blue Blook. Its fourth or fifth released exam i think.</p>
<p>It says the answer is 5. You’re doing it correctly.</p>
<p>For some reason my book says 6. I dunno how a misprint like that happens…thats really weird. </p>
<p>Well at least im not an idiot. Couldn’t believe i missed such an easy question.</p>