How do you approach problems that are seemingly ridiculous?

Test Preparation SAT Preparation
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<p>For example, I encountered this problem and was flabbergasted:
What is the remainder when 2^400 is divided by 10?
A. 0
B. 2
C. 4
D. 6
E. 8</p>

<p>Now, my book tells me the answer is C or D (Don't ask; there's a glitch in the answer key that sometimes states the answer as 18. C, then follows it up with 18. D). However, I have essentially no clue how to get this. I have a vague idea on how to solve this, but I'd rather someone explain this to me.</p>

<p>On a lesser scale, another example of these "ridiculous" problems is thus:</p>

<p>The greatest integer in a list of 101 consecutive integers is 111. What is the median of the list?</p>

<p>This problem isn't bad, in my opinion, after writing the list counting backwards, I quickly saw that the nth term was n + 10, and the median of 101 would be 50.5. I proceeded to add 10 to this, and arrived at 60.5 as my answer. I still have this nagging feeling that I got that problem wrong and it could be 61, but nonetheless, my Princeton Review book tells me the answer is 56. Again, without an explanation, I am lost. Would anyone care to shed light on these problems? I really despise Sequence/Growth problems.</p>

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<p>Question your book. The series median is simple to compute with Excel to check your answer vs. the book answer. (Use the fill function to develop the list of integers.) For 2^400, look for a pattern.</p>

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<p>look for patterns.</p>

<p>2^1 ends in 2
^2 4
^3 8
^4 6
^5 2
^6 4
…</p>

<p>see, it’s a 2 4 8 6 2 4 8 6…</p>

<p>pretty much it’s asking you what the 1’s place is in 2^400.</p>

<p>when our patter is 2 4 8 6, in which repeats itself once every 4 terms. Divide 400 by 4 and you dont get a remainder.</p>

<p>Meaning that you can repeat 2 4 8 6, 100 times, and you still end up at the END of the first term. Which is a 2 4 8 [6]. 6, Answer C.</p>

<p>-
Try this simpler version if you dont understand my justification</p>

<p>What is the remainder of 2^5 if it’s divided by 10?</p>

<p>Forget that we know it’s 32</p>

<p>Knowing it’s the 5th power, 5/4 mod 1 (this is another way to express remainder)</p>

<p>So we know that we repeat the 2 4 8 6 pattern once, and go over 1, in which gives up 2 4 8 6 [2]. And 32 does end in 2!</p>

<p>So when we get 400, which has no remainder. We basically go over 2 4 8 6 a hundred times and have no remainder, and we land on the last number, 6.</p>

<p>That’s much easier to explain with a piece of paper and my mouth.</p>

<p>Edit: I just read your stuff is from PR, which in my opinion is for chuckles. Some of the grammar problems I have done on there weren’t even English. I dont have a fond opinion of it.</p>

4

<p>um, he listed 6 as answer D. That’s why I said question the book.</p>

5

<p>Oh, typo on my behalf. But 6 seems like the right answer, if i didnt misread anything</p>

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<p>smash20, also double check the median of 101–that’s an odd number.</p>

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<p>I am too lazy to think about stats right now but I did it in excel and it yielded 61</p>

<p>101 consecutive int. with 111 being the largest is an list from 11 to 111 (I kept thinking 10, need more sleep)</p>

<p>Your second guess was correct. But remember it’s an odd number so it yields a whole number median.</p>

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<p>Kidwithshirt:</p>

<p>Thanks a bunch, I understood your first explanation; I didn’t even think to look for patterns. And, if it changes anything, this isn’t a book PR puts out for the masses, you can only receive it if you take the course (it’s called Beat the SAT), at least that’s what my instructor told me. I just figured I might as well go through it, since it can’t hurt me and would be a waste of 425 pages otherwise. I’m planning to go through that, then the Blue Book, then Barron’s 2400, then if needed, Gruber’s Complete SAT Math Workbook. If doubt I’ll be able to go through all that for the May SAT, but I’ll leave the remainder over for October/November. I do agree, though, that PR is riddled with mistakes. It’s a pain, but what can I do? Only book that seems not to have them is the Blue Book.</p>

<p>As for the second problem, I see that it is 61; I should have just taken the average of the first and last terms, determining them by the formula n+10, where n is any number. I made the same mistake of thinking that 10 was the first term (I subtracted 101 from 111), which is what threw me off.</p>

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<p>Heh. When it’s a large number, which normally cannot displayed by your calculator, it’s usually a problem like this in which you find patterns by just looking at the end numbers.</p>

<p>I always get it mixed up because I always think that if it has no remainders, it should start on the first term since I consider having no remainder yields a new sequence of numbers (i try to make sense, but I dont think I did)</p>

<p>I just thought of it differently after completing your problem. Just think that if it has no remainder, it’s a remainder of, in your case, 4 (even though it’s the divisor, but just think it that way) so it’s the last number in the sequence.</p>

<p>For the stat problem, I encountered almost the exact same on on the DEC SAT except it was around 10 to 40 range. I wasn’t thinking straight that day so I wrote out all the numbers and counted in pairs. I realized that averaging would do the trick since they were consecutive. </p>

<p>I recommend studying for the SAT II math level 2 like someone else recommended, it would help a ton with the basics.</p>

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<p>Yeah the cycle way is good. If you’re confused by the 400th power, make the problem simpler - what if the 400 were a 4? Then 2^4 = 16, and the last digit is 6. Adding the extra 0’s to the 4 doesn’t change the problem.</p>

<p>For the median one: n+10 describes the sequence. Then the first term is 11 and the 101st is 111. To get the median, there must be an equal number of numbers of both sides of it. It’s not too hard to figure out that there must be 50 on both sides (111-11 = 100, divide by 2 = 50). Then 11+50 = 61 or 111-50=61. This answer makes sense as the median of a list with an odd number of numbers is a whole number.</p>

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<p>just get rid of the 400 and use 2^4 instead…</p>

<p>2^4=16 16/10 has a remainder of 6</p>

<p>when you are faced with problems like this, you can almost always reduce or make the question easier. obviously the SAT doesn’t expect anyone to be able to do 2^400</p>

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<p>Usually these sorts of ‘ridiculous’ problems involve patterns or sequences, but I recently encountered this geometry problem that fazed me because it didn’t seem to have all necessary information:</p>

<p><a href=“http://i44.■■■■■■■.com/27yqwoz.jpg[/url]”>http://i44.■■■■■■■.com/27yqwoz.jpg&lt;/a&gt;&lt;/p&gt;

<p>If triangle ABCD shown above has an area of 24, what is the probability that a randomly chosen point that lies in ABCD will be in the shaded region?
A. 1/3
B. 1/2
C. 2/3
D. 1
E. It cannot be determined from the information given.</p>

<p>I guessed 1/2, which is correct, according to my book. I chose it for two reasons:

  1. The answer is never “It cannot be determined from the information given.”
  2. I just imagined a string connected to points A and B, and thought that since it could cut the triangle in half if it touched point B or C, it would be able to do that no matter which point on BC it touched.</p>

<p>However, I don’t really trust my logic because I didn’t incorporate the area. Could anyone explain how to solve this the way College Board wants you to?</p>

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<p>Uh, think about it. The height of the triangle is the height of the quad, and the base of the triangle is the base of the quad. </p>

<p>Area of quad = l<em>w = 24
Area of tri = 1/2(l</em>w) = 12 </p>

<p>Thus, it’s 12/24 -> 1/2.</p>

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<p>smash20: triangle ABCD???</p>

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<p>He means rectangle…</p>

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<p>or he meant triangle whatever it was in which case your numbers would be 24 and 48 rather than 12 and 24, it’s all good.</p>

17

<p>Wow, I can’t believe I overlooked that. My logic about the strings is still valid, though, but now I can see why.
How about this following question, it’s really baffling me:</p>

<p>On a cruise ship, there are 120 passengers for every officer, 60 passengers for every steward, and 100 passengers for every chef. If the crew is made up only of officers, stewards, and chefs, and if the crew totals 84 persons, how many passengers are on the ship?</p>

<p>The answer is apparently 2400, but I don’t know how that works. Wouldn’t the least number of passengers be 5040, if the crew was made up of all stewards? This is an open-ended question, btw.</p>

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<p>Well, I got 2400 like this:</p>

<p>Least Common Multiple of 60,120, and 100 is 600. Since the passengers don’t only belong to one steward, officer, or chef, they can overlap - the officer shares her 120 with two stewards and 1.2 chefs. </p>

<p>So, for those 600 passengers there are 5 officers (120<em>5=600), 10 stewards (60</em>10=600), and 6 chefs (100*6=600). So, adding up those numbers (5+10+6) iequals 21 total crew for those 600 passengers.</p>

<p>Since 21<em>4=84, the number given in the problem, you multiple 600</em>4 to get 2400.</p>

<p>Does that make sense?</p>

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<p>Yeah, it does. I completely misread the problem and didn’t realize passengers could be shared. Thanks a bunch.</p>

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<p>All right, I got a couple fairly hard problems that I need help on; any explanations would be appreciated.</p>

<ol>
<li>5 people, Marie, Pauli, Riani, Shivani, and Terrance are riding in a car during a cross-country road trip. There are 5 eats available, and any of the passengers can sit in any of the passenger seats but not all of them can drive. If there are 72 different ways that Marie, Pauli, Riani, Shivani, and Terrance can sit in the car, how many of them cannot be drivers?
The book says the answer is 3, but I got 2 as my answer. If we go from most restricted position (driver’s seat) to least (passenger’s seat), I can compute this if 2 cannot drive: (3 possibilities)(4 possibilities)(3)(2)(1)=72 This seems to work out, but I am supposedly wrong.</li>
</ol>

<p>As for the next one, I really don’t have a clue:
2. (x + a)(x + b) = x^2 + cx + d
In the equation above, a, b, c, and d are all integer constants. If d is prime and a<b, all of the following are equal to 1 EXCEPT
(A) a
(B) b/d
(C) c-d
(D) c/(a+b)
(E) d-b
The book states that the answer is E.</p>