<p>
</p>
<p>I’ve been thinking about this all day. Is this a real value? or simplification?</p>
<p>There’s a really silly way to simplify this. It’s algebra II math, to be honest. You want me to PM you a hint?</p>
<p>I think I’m going to have one more day to think about it… I’ll PM you, if I need a hint… 
Thanks.</p>
<p>
</p>
<p>Actually, the question I answered is a nice puzzle, but has no relevance whatsoever to the SAT for a number of reasons.</p>
<p>If you like sequences, here are two:</p>
<p>A. 5 ? 1,729 635,318,657 </p>
<p>B. 1000 999 972 954 936 918 900 ?</p>
<p>@xiggi</p>
<p>Yeah,I know. They are more like fun puzzles than anything else. I was joking, anyway. You can obviously do whatever you please. If you want a true puzzle, solve the one Jeffery is doing. It’s the kind of question that makes you feel silly for not realizing the solution. Did you play this game called Professor Layton? I swear that some of the puzzles in that game can be used here. Maybe I’ll look for a few questions in that game…</p>
<p>@xiggi</p>
<p>I hate sequences. </p>
<p>For B, I believe that the rule is simply descending. The thing about abstract sequences is that you can find an infinite number of mathematical rules that work for all the sequence (by simply concocting weird rules), but those rules can easily fall apart at the last term. Also, instead of trying to prove something right, isn’t a better idea to prove something wrong in such a nebulous situation (like anti-knowledge or something)? I could say something like 8+1=9 in 918 and 3+6=9 in 936, but that’s really just trying to prove logic that might not exist. Imma keep this simple and say that the rule for the sequence is descending. If I’m wrong, please give me a hint or something to work with. </p>
<p>Did you forget a comma in A? The formatting is confusing me.</p>
<p>I think I’m thinking too hard about this stuff. I’ll do something dumb now, like watch cartoons or read junk.</p>
<p>The sequence is descending and it is not nebulous. Very straightforward. </p>
<p>No comma was forgotten in the first one.</p>
<p>First, 2^1.5 = 2^(3/2)= 8^.5= 2(2^.5). So we rewrite this equation as 2^.5-(3-2(2^.5))^.5. </p>
<p>Now, we want to get rid of the radical on the right, so we set -(3-2(2^.5))^.5 = a+b(c^.5), (I’m not sure why, but in all problems like that, a double radical is equal to a single radical in another form). Squaring both sides yields 3-2(2^.5)=a^2+2ab(c^.5)+b^2. Because c is the only term under the radical, it must equal 2. Now we match like terms to get 2 equations. a^2+cb^2=3 and 2ab=-2. By the second equation, ab=-1 and b=(-1/a). Substituting that and c=2 into the first equation yields a^2+(2/a^2)=3. Combining the fraction yields (a^4+2)/(a^2)=3 and simplifying yields a^4-3a^2+2=0. By observation, we see that the sum of the coefficients = 0, so a= 1,-1. Because we originally have -(3-2(2^.5)), a must equal 1, since -(3-2(2^.5)) is obviously negative and in a+b(c^.5), a must be > than b to yield a negative value as well.</p>
<p>So substituting a=1 into ab=-1 yields b=-1. So a=1. b=-1, c=2 and -(3-2(2^.5))=1-2^.5 and our desired answer is 2^.5+1-2^.5= 1</p>
<p>@xiggi</p>
<p>B) Add all the digits of the number and subtract the sum from the number itself. The solution is 891. I guess I was overthinking it.</p>
<p>@cortana431</p>
<p>Interesting solution. I never thought about it in that way. I’d like to share my not-so-sophisticated solution: [View</a> image: Radical question](<a href=“http://postimage.org/image/73m5rvl6t/]View”>http://postimage.org/image/73m5rvl6t/) :p.</p>
<p>@cardgames</p>
<p>I think your solution is actually quite sophisticated. When I did the problem yesterday it was not immediately evident to me that rewriting 3=2+1 would give a perfect square under the radical. BUT I did know that I wanted a perfect square under the radical AND that sqrt(2) needed to appear (since the question seems to imply that we want a solution without any roots). So I simply attempted to solve the following equation:</p>
<p>3-2sqrt(2)=(sqrt(2)+x)^2 = 2+2xsqrt(2)+x^2</p>
<p>Setting the constants equal and the coefficients of sqrt(2) equal we get:
2x=-2 and x^2=1. This system has the unique solution x=-1. </p>
<p>Thus, 3-2sqrt(2)=(sqrt(2)-1)^2, and it follows that the original expression simplifies to 1.</p>
<p>It looks to me like you rewrote 3 as 2+1 just by using trial and error (I don’t necessarily mean this literally - upon reflection I should have seen this). I’m curious if you found this using a more formal method that can be generalized to similar problems with numbers that are not as nice (my method generalizes quite easily).</p>
<p>Ok. Let me give some number theory problems. I’m going to start off a bit easier, and work up to something more difficult and substantial. Let me start by reminding you of a definition:</p>
<p>An integer n is even (or divisible by 2) if n=2k for some integer k.
More generally, an integer n is divisible by d if n=dk for some integer k.</p>
<p>Questions:
(1) Show that the product of an even integer and any other integer is even.
(2) Show that the product of 2 consective integers is even.
(3) Show that the product of 3 consecutive integers is divisible by 6.</p>
<p>@Drsteve
I didn’t really follow a formal method (I often don’t). I also knew that I needed a perfect square pretty fast. Therefore, I thought 3-2sqrt(2) = a^2 +2ab +b^2. It’s obvious that ab=-sqrt(2), and a^2+b^2=3. You do a little algebra and realize that 1 is b^2–the rest is obvious. That’s all there’s to rewriting the 3.</p>
<p>By the way, thanks for the recommendation. I was talking about abstract algebra.</p>
<p>^^
Does that mean that the product of 4 consecutive integers is divisible by 24. ;)</p>
<p>PS It is.</p>
<p>
</p>
<p>For that problem, in addition to using the same method as CG (fwiw, it comes up in GMAT solution tricks), I also like to use Google’s bar. This is what shows up. </p>
<p>(2^.5) - ((3 - (2^1.5))^.5) = 1</p>
<p>Of course, it IS a calculator. ;(</p>
<p>Does that mean that the product of n consecutive integers is divisible by n! ? (use induction if you’re feeling up to the task)</p>
<p>I’ll leave those to the guys who aren’t familiar with number theory.</p>
<p>@xiggi</p>
<p>Are you sure that sequence(a) isn’t missing a comma? It has been owning me for quite a while. Help?</p>
<p>The numbers are correct.</p>
<p>5
?
1,729
635,318,657 </p>
<p>Check some of the properties of the number 1729. It is a special number. For instance, the sum of its digits is 19 and the product of 19 and 91 is 1729. But this sequence has to do with squares. </p>
<p>For a hint, google the term taxicab. :)</p>
<p>@Xiggi and Cardgames</p>
<p>Those were going to be questions (4) and (5)</p>
<p>(4) Show that the product of 4 consecutive integers is divisible by 24.
(5) Show that the product of n consecutive integers is divisible by n!</p>
<p>@Cardgames </p>
<p>When I teach graduate students group theory I use “A Course in Group Theory” by John F. Humphreys. It’s not too advanced for a graduate text. I also have both student and instructor notes that are pretty self-contained - the instructor notes have solutions to some pretty standard problems that you might have trouble finding elsewhere. I’d be happy to e-mail them to you if you’re interested.</p>
<p>If you want something more basic, I would guess that Hungerford’s book is good, but I admit that it’s been a while since I’ve checked out any undergraduate texts in abstract algebra.</p>
<p>Another sequence question…definitely NOT sat material…and kind of obnoxious but one of my favorites nonetheless. (Apologies if I have posted this in the past – I don’t remember.) Here goes:</p>
<p>1, 11, 21, 1112, 3112, 211213, …</p>
<p>Find the next term. And find the limit of the sequence.</p>
<p>If you are stuck, I can give a hint that you wouldn’t think would be helpful but turns out to be very helpful.</p>