mxplusb

1. r^2-r kamakshi is correct, and that approach is very simple and elegant. I was initially thinking of it sueka's way: the whole rectangle's area is r(r+1), then you take out one row and one column, and add one for the square you took out twice. That gives you: r(r+1) - (r+1) - r +1 = r(r+1) - 2r = r^2+r-2r = r^2 - r. When you take off the outermost row and the outermost column, you're essentially taking one square out twice, the same way as you do when you take out an inner row and an inner column. 2. 1/8, or .125 It seems like everyone's good on this, but just in case: 1,000 is evenly divisible by 8, so out of the first 1,000 positive integers exactly one out of eight, or .125, are divisible by 8. Of course every number that is divisible by 8 is also divisible by 2, so that doesn't change anything. There is a certain amount of unnecessary information in these questions just put there to throw people off. 3. 1/6 This is a combinations question. Combinations = order matters, permutations = order doesn't matter. So with combinations, "2 and 6" and "6 and 2" are the same. (If you wanted to use permutations you could, but then the total number of possibilities would be 4P2=12 rather than 4C2=6.) 4. 100 I'm assuming

Agreed. Substituting real numbers is often a good approach -- don't do it in math class, except to check your work, but it works well on standardized tests.

Thank you. I try.

thepillow's explanation looks good to me. Thank you for that. I find that probability and combination/permutation problems are the most confusing problems on this sort of test. There are many different rules that can be applied, and it is often hard to figure out which one fits best. Personally I hate having to rely on an answer sheet to tell if I got the right or wrong answer -- I'd much rather be able to double-check myself. Of course on standardized tests there isn't always time. But when you're studying there is -- either you can allow yourself all the time you need to get the problem right (remembering that practicing doing a problem wrong is worse than not practicing it at all) or you can go through a practice test once while timing yourself, and then go back to the problems you're not sure of and take your time on the second untimed round, before looking up the answers. Permutation and combination formulas are shortcuts. Many problems you can test out long-hand, by actually writing down all the possibilities, and on the ones you can't you can often make a smaller version of the same problem and test that one. This way you don't have to take the strategies on trust. You can use them because you know they work, you've tested it yourself. I was thinking about this, because I've noticed the way I do test problems is often different than the way a textbook would say to do them. Textbooks tend to focus on memorizing formulas and strategies, then figuring out which one to use. I tend to not lean on formulas unless I'm very sure about them, either because I've used them so much that they're second nature or because I know how to derive them. I also rely on estimation and common sense a lot -- not relevant for this problem, but some problems will have one or a few answers that are clearly impossible. So I can cross those off without even checking.

It is amazing how quickly math knowledge can fade, if you didn't really drill it in when you first learned it. I would recommend spending a LOT of time on practice problems. The more you practice, the more it will be automatic on the test and the less time you will spend second-guessing yourself. Personally -- and your mileage may vary -- I almost never "study" formulas or attempt to memorize them. I do problems. I believe that if you do a lot of problems, with the formulas written down somewhere you can check them easily, you will probably not have to deliberately memorize them. Instead you will automatically learn the formulas by heart just from using them so much. At any rate, I've gotten A's in quite a lot of math classes with that approach.

Ah! Thank you for looking that up.

That's beautiful. I never would have looked at it that way. Anyway, thepillow, there's a way that's slightly faster than your first approach, I think. You take all the digits combinations -- 0005, 0014, 0023, 0122, 0113, and 1112. For the first and last, the ways to arrange the digits are 4!/3! (the 3! is because the three 0's or three 1's are interchangeable). For the others, it's 4!/2!. Adds up to 56. In this case it's okay for the numbers to start with a zero because that just represents a less-than-four-digit number, which is allowed. And I would have gotten the correct answer very quickly, if I hadn't forgotten about the 0122 combinations. Drat.

I will not give a rating because I am not sure of the rubric used. But my first impressions are: solid grammar and use of vocabulary; the arguments are clearly stated and there is a good flow that is easily followed. Also the counter-arguement is acknowledged appropriately. Some potential improvements: The fourth paragraph, which addresses counterarguments, would be better placed either as the second paragraph or as next to last, so it doesn't break up the flow of the arguments that you are making. The second paragraph starts out very clear and strong, but the last sentence is confusing, in part because the word "position" is used too many times. (Perhaps "field" would be an appropriate substitute for the middle one.) Consider dedicating a full sentence in the opening paragraph to each of your three points, rather than combining two into one sentence. The word "initially" in the last sentence is, I think, inappropriate: you're talking a problem that is likely to get worse over time, rather than one that is likely to be most acute immediately after graduation. "Actually" or "in fact" would fit better. Also consider whether your main premise is better stated by your first sentence or your final one -- they are not quite the same statement. The first statement leaves unclear whether the availability of jobs should be considered at all. If your premise is that it should only be one factor of several, make it clear in the first paragraph. It is also not clear whether you think that such a factor is likely to "bolster the prospect of future employment" (but other concerns like quality of life are important too) or that the possibility of a changing job market cancels out that potential advantage, making that perception largely illusory. It seems like you're trying to use as many intermediate to advanced vocabulary words as possible; generally you use them accurately, but you may wish to use the principle that a little of that goes a long way, and tone it down a bit. (Then again that may just be my own bias speaking: I work in education, where I typically have to state concepts in a way that 10-year-olds will understand. Of course GRE applications require a different emphasis.) I think your essay has a lot going for it as it stands, and that those changes could make it an outstanding essay. Again, you've already got good grammar, good spelling, a good vocabulary, and a mostly clear, readable structure.

I like the tone of this essay -- it sounds like you're writing about something you care a lot about personally. The structure is solid: your thesis is clearly stated at the beginning and repeated in the conclusion, and each example is given an equally-weighted paragraph. The writing is also concise, without unnecessary phrases or sentences.

... unless you can use each digit more than once. In which case I will have to throw in my support for 60. OK, clearly I misunderstood the problem.

If I am understanding the question correctly, there are five digits to work with: 0,2,5,7,9 There are 5P3 = 5*4*3 = 60 ways to arrange these five digits into groups of three. One-fifth of these will start with a 0, so they are not proper three digit numbers. That leaves 48 combinations. Two-fifths of the original 60 combinations end in a 0 or 2, making the number even. Three of these combinations -- 052, 072 and 092 -- have already been eliminated for starting with 0. Take out two fifths of 60, or 24, from 48, then add back the 3 that got double-counted. That leaves 27. (I was initially skeptical of this answer, since it's not given as an option, but I have checked it in multiple ways including in the end writing out all possible combinations, so I am quite certain it is correct. Unless the question was copied wrong or I misunderstood.)

... can you come up with an example of how this might come up in a question? I feel like this question should be obvious -- but suddenly I'm unsure whether to say "the prime factorization of 1 is 1" or "1 doesn't have a prime factorization, because it's just different that way." Anyone else have any thoughts?

"further more there are only two types of right triangles one is 30-60-90 and the other is 45-45- 90" Completely wrong. As a quick example, take a piece of paper. Each corner is a right angle. Fold one corner over. Now you have a right triangle, with the folded edge as the hypotenuse. Are there only two triangles you can fold this way? Hardly.

Pythagorean equation: a^2+b^2=c^2 All the outer triangles are congruent (this is easily provable by geometry if you're not willing to just eyeball it) with one outer side 2 and one outer side 1. So the hypotenuse -- which is also the side of the inner square -- is root(2^2+1^2) = root (5) The area of a square is the square of the side. So, (root(5))^2 So the area of the square is 5.

It is not a rule, just an observation. 32=8*2*2 Therefore, any number that is divisible by 32 must also be divisible by 8 so if it is not divisible by 8 it is not divisible by 32. Also dividing by 2 twice and then checking for divisibility by 8 is exactly equivalent to checking for divisibility by 32.