<p>I didn't take the May SAT, but I'm curious about the solution to this problem:</p>
<p>[2x^n+1+X^N+2=X^n+3 find x^3]</p>
<p>I dont know the options, but how would one go about solving this? Thanks alot!</p>
<p>substitute 0 for n.</p>
<p>I believe it was said that n could be any integer. Anyway, one would plug 0 in for n. Giving: 2x+x^2</p>
<p>Thanks!
10char</p>
<p>oh God…it’s not that I got it wrong, it’s just i wasted 2 minutes on this problem…
I could have substituted 0, but I took the long way and yea… I gotto learn to do things the easy way to improve my score</p>
<p>It should be solved like this:</p>
<pre><code> 2x^(n+1) + x^(n+2) = x^(n+3)
</code></pre>
<p><=> x^n * (2x + x^2) = x^n * x^3
<=> x^n * (2x + x^2 - x^3) = 0
<=> x^n = 0 (I)
or (2x + x^2 - x^3) = 0 (II)</p>
<p>(II) => x^3 = 2x + x^2 </p>
<p>That’s it.</p>
<p>yea thats what i did, thats why it took me so long…should have just substituted</p>
<p>What did you guys put for that greatest 4 digit number that follows the 3 guidelines problem? I had 8521 but I’m not sure if it’s right</p>
<p>8521 is right</p>
<p>8521 is correct.</p>
<p>Anyone remember if the x^3 the answer was B?</p>
<p>Is there a systematic way to go about finding 8521?
Silly me, rushing to finish the section, thought it said three-digit number. I think only I could do something that stupid.</p>
<p>@reginaphalange</p>
<p>OMG I DID THE SAME THING… your not alone =(
I got a 3 digit number that satisfyed the conditions… this is how I kill my score</p>
<p>Ahh, so glad I’m not the only one! That’s exactly what I did.
I was discussing it with people later “What did you get for the three-digit number that was the last question?” “You mean the four digit?” “No, it was three” <em>asks someone else</em> “Uh, it was four…”
I’m intelligent enough to get SAT scores around the 95th percentile, but not intelligent enough to read a question. Gotta love it.</p>
<p>The only way it put in numbers i think. And make the last three equal the first. I dont think there is a way except guessing and checking</p>
<p>@regina</p>
<ol>
<li><p>Firstly, you had to know that none of the digits could be 0 because, if one was, the product of the four digits would be 0.</p></li>
<li><p>Secondly, you had to realize that the product of the four digits was divisible by 10. To achieve this, you’d need either a 2 and 5, or a 4 and 5 to make sure the sum of the three smallest digits was <=9. Since 4+5=9 and you still have another digit to add which cannot be 0, you must use 2 and 5.</p></li>
<li><p>To preserve the property of being divisible by 10, the 2 and 5 would need to be paired with either a 1 or 2, because any integer greater than two, say, 3, would give you 2+5+3=10, and 10 cannot be a single digit, obviously. You also could not use 2, because that would result in the 2s being repeated, so you would need to use a 1.</p></li>
<li><p>You now have 2, 5, and 1. Add them up to get 8. 8521</p></li>
</ol>
<p>Thank you so much, enmar. Those ones always trip me up, when they’re the last on the section and I’m rushing.</p>