A few difficult ones...

Test Preparation SAT Preparation
1

<p>If 2007^2007 is multiplied out, the units’ digit in the final product is:</p>

<p>Thirty-one books are arranged from left to right in order of increasing prices. The
price of each books differs by $2 from that of each adjacent book. For the price of
the book at the extreme right a customer can buy the middle book and an adjacent
one. Then:
(a) The adjacent book referred to is at the left of the middle book.
(b) The middle book sells for $36.
(c) The cheapest book sells for $4.
(d) The most expensive book sells for $64.
(e) None of these is correct.</p>

<p>I might add more in this thread.</p>

<p>I would be grateful if you could provide the working as well.</p>

<p>Thanks for the help!</p>

2

<p>hi :)</p>

<p>for the first one, you just keep multiplying the last digit by 7 until you see a pattern in the last digit. here, the last digit is 7, so i’ll give you the pattern for the last digit. 7^1=7; 7^1 * 7=7^2=<em>9; 7^2 * 7=7^3=</em><em>3; 7^3 * 7=7^4=</em><em>1; 7^4 * 7=7^5=</em>__7. Since 7^5 and 7^1 have the same last digit (7), the pattern repeats every four exponents. 2007/4 gives you a remainder of 3, which means that the last digit will be the same as 7^3, which has a last digit of 3. </p>

<p>ask me more if it doesn’t make sense. it’ll take me a while to type up an answer for the 2nd question :D</p>

3

<p>just do 2007^2007 on the calculator :D</p>

<p>oh wait it overflows, then I would think that since 2007^2 ends in 49 and 2007^3 ends in 343, then 2007^7 must end in a 3, so i would guess that 2007^2007 ends in a 3</p>

<p>i could easily be wrong, but I doubt that you would need to know that one for the SATs because if a calculator cant even do it, they wouldnt expect a high school student to be able to deduce it</p>

<p>for the 2nd one, i would test them, starting with D(you should start with C but i felt like D :D)</p>

<p>if 64 is the most expensive book, then the middle book is the 16th book(well its always the 16th book)</p>

<p>that means that the 16th book is 30 dollars cheaper than the 31st book, and if the middle book is 34, you wont be able to buy a 34 and a 32 with the 64</p>

<p>so D is wrong</p>

<p>C: if 4 is the cheapest, then the 16th book is 30 dollars more expensive, and the 31st is 30 dollars more expensive than the 16th</p>

<p>so if the middle is 34, well you could stop right there because its a repeat of D</p>

<p>B: if the middle book is 36, then the 31st book must be 66, and you cant buy 36 and 34 with 66</p>

<p>A: if the middle book is 32, the last book is 62, and the book immediately to the left of the middle book (the 15th) is 30</p>

<p>so yes you can buy that</p>

<p>the answer is A unless i screwed up at the beginning</p>

<p>EDIT: blast, i was beaten by tiny mermaid. his or her method is also more efficient. although we reached the same answer, I reached it with a method that is basically fancy guessing mixed in with pattern recognition.</p>

4

<p>me again,</p>

<p>the formula for each book here is x+2(n-1), where x is the first book’s price and n is the book’s number. for example, if the first book’s price is $2, its formula would be 2+2(1-1)=$2. The second would be 4, etc. The last book’s (#31) price is equal to the middle (#16) and the adjacent (#15 or #17, we don’t know which). let’s test the cases.</p>

<p>(a) if the adjacent book is to the left of the middle, then it would #15. so, x+2(15-1) + x+2(16-1) = x+2(31-1). this would come out as 2x+58=x+60, or x=2. This is possible, so (a) is correct. If the book had been on the right, it would’ve been x+2(17-1), and x would have had a negative value, which is impossible since the first book would then have a negative price. If you want to test the other options, just plug in numbers into those two options for each one and test (although none of them should work). </p>

<p>okay that was so wordy. yeah, ask me if this one doesn’t make sense too.</p>

5

<p>haha sorry antonioray XD</p>

6

<p>again, tinymermaid beats me with a more efficient answer</p>

<p>or is it merely different methods?</p>

<p>could it be that tinymermaid is more intelligent?</p>

<p>hmmmmmmmmmmmmmmm</p>

<p>EDIT: well she(or he) did start testing with the correct answer, and I indiscriminately chose an answer in the middle of the pack</p>

7

<p>nah it’s probably just that i’m more intelligent</p>

<p>^^i’m so modest^^</p>