<p>The sum for the 3 days is 75(3) = 225. The sum for the 5 days is 100(5) = 500. So the sum for Thursday and Friday is 500 - 225 = 275.</p>
<p>Since I haven’t gotten any response to the second problem I wrote, I’ll repost it.</p>
<p>Level 5 Counting: </p>
<p>The diagonals of a square divide its interior into 4 congruent, non-overlapping, triangles. Given the three colors red, yellow, and green, how many ways can a color be assigned to each of these triangles in such a way that any two triangles sharing a side do not have the same color?</p>
<p>this is an awesome thread</p>
<p>@music202</p>
<p>C can not be true. If 6 is in p, then it is also in q, as given in the problem statement (“every number in p is also in q”), which contradicts the answer choice. The other answer choices are perfectly possible. I can go through them to show why if you like</p>
<p>@RandomHSer Thanks! I understand it now.</p>
<p>When a certain positive number “n” is divided by 6 the remainder is 3. When “n” is divided by 7, the remainder is 4. What is the least possible value of n?</p>
<p>I put down the answer as n = 3, but it was incorrect. Could anyone explain why?</p>
<p>Is it because the dividend cannot be smaller than the divisor if you’re trying to yield a remainder (i.e : 3/6, 3/7)?</p>
<p>The answer is: 39.</p>
<p>Any help would be appreciated,</p>
<p>When 3 is divided by 7 the remainder is 3. Place 3 coins in front of you. Now try to group them 7 at a time. How many groups do you have? How many are left. That’s right - there are 0 groups and 3 left over. </p>
<p>Thanks for the help @Drsteve.</p>
<p>these types of questions always seem to get me…i really need a dumbed down or simplification.</p>
<p>"For all numbers a and b, let a 
b be defined by a b = ab + a + b. For all numbers x, y, and z, which of the following must be true?</p>
<p>I. x y = y x
II. (x-1) (x+1) = (x x) - 1
III. x ( y+z) = (x y) + (x z)</p>
<p>a. I only
b. II only
c. III only
d. I and II only
e. I, II, and III. </p>
<p>what does actually represent? when it says x , what does that stand for? or just (x-1) (x+1) , what is that equal to? </p>
<p>For the counting problem, it’s not clear if all 3 colors have to be used. I’m guessing this is not a College Board question. Also this is more difficult than any SAT problem I’ve seen. I’m going to assume that all 3 colors kit be used. </p>
<p>Either the left and right triangle or top and bottom triangle must have the same color. So assume the left and right triangle have the same color. Then we have 3 choices for the left, 2 for the top and 1 for the bottom. So there are 3<em>2</em>1 = 6 altogether. Multiply by 2 for the other case. So the answer is 12. </p>
<p>@alphabetsoups doesn’t “represent” anything other than an operation some SAT writer made up for a question. x:slight_smile:on its doesn’t mean anything. As you can see, the operand was only defined to take two inputs. (x-1)(x+1) means that we plug in x-1 for the a in the definition of and likewise x+1 for b.</p>
<p>I is true.x☻y=xy+x+y and y☻x= yx+y+x. These two expressions are equal.</p>
<p>II is true. (x-1) (x+1)= (x-1)(x+1)+(x-1)+(x+1)=x^2-1+2x. (x:slight_smile:x)-1= x^2+x+x-1=x^1-1+2x. So these two expressions are equal.</p>
<p>III is false. x☻(y+z)=x(y+z)+x+(y+z)=xy+xz+x+y+z but (x☻y) + (x☻ z)=xy+x+y+xz+x+z=xy+xz+2x+y+z so they’re not equal.</p>
<p>“Either the left and right triangle or top and bottom triangle must have the same color. So assume the left and right triangle have the same color. Then we have 3 choices for the left, 2 for the top and 1 for the bottom. So there are 3<em>2</em>1 = 6 altogether. Multiply by 2 for the other case. So the answer is 12.”</p>
<p>I and II only are true- D.</p>
<p>@DrSteve</p>
<p>I didn’t intend for it to mean that all colors must be used, so my intended answer was 18 (adding the six cases where both opposite pairs of triangles have the same color), but you’re right that it’s an ambiguous question. I wrote this question and the last; they’re not College Board questions (were all the questions you posted CB questions?). I’ll stop.</p>
<p>thank you, i think i’m starting to get it. the math section is going to be the death of me, lol. </p>
<p>@RandomHSer</p>
<p>I didn’t realize you posted that or I would have waited for students to answer. I actually like the problem a lot (as long as you clarify that all colors do not need to be used). I would classify this as a “Level 6” problem. It requires just a bit too much skill to be an actual SAT problem (at least in its present form), but it could possibly be an SAT problem if the SAT were just a little harder.</p>
<p>I actually created a thread a while back on “Level 6” problems, but it sort of got hijacked by lots of people posting questions that were much harder than Level 6 problems. I don’t mind this. If I start a thread and it takes on a life of its own with less and less involvement from me, that’s a good thing. Depending on how busy I am I don’t always have the time to keep these threads going. So the last thing I want to do is discourage anyone from keeping these threads alive. So if I am not updating one of the threads I started, please feel free to try to save it. </p>
<p>I hope that the threads I start provide a lot of value for students that may not be able to afford tutors/courses or even books. So let’s all keep them going.</p>
<p>@DrSteve</p>
<p>I always have trouble gauging the difficulty of counting problems. On tests like the SAT and harder tests like AMC/AIME, combinatorics is always my best subject by far so I tend to underestimate its difficulty (and the opposite is true for geometry. I think I overcompensated for this when I gave my first problem a 3/4 difficulty).</p>
<p>I’ll look into that thread. I enjoy a good challenge.</p>
<p>A Level 5 SAT counting problem will usually involve the counting principle with a small twist. For example arranging 5 things such that 1 or 2 of them can’t be placed in a specific position. Another possibility is that the counting principle might be very tricky to apply, but the question can be solved by creating a careful list.</p>
<p>I also have a thread of “Challenge Questions” out there which are mostly of a theoretical nature - much harder than SAT problems.</p>
<p>@alphabetsoups I noticed that your question is from the Jan 2006 SAT sec 2, #20. Do you know how to do #19 in the same section?</p>
<p>if s,t,u, and v are the coordinates of the indicates points on the number line above, which is the greatest? The number line shows that s is B/W -3 and -2, t B/W -1 and 0,u B/W 0 and 1, and v B/W2 and 3.
absolute value for all choices
A) Is+tI
B) Is+vI
C) Is-tI
D) Is-vI
E) Is+uI</p>
<p>Thanks!</p>
<p>@RandomHSer - IMHO its fine to post non-CB questions, but it would help to identify them as such</p>
<p>@CHD2013 Point noted. I will do that in the future if I post any more of my own problems.</p>
<p>@music202 </p>
<p>|a-b| is the distance between the points a and b on the number line and so |a+b|=|a-(-b)| is the distance between a and -b (or equivalently, between -a and b). We see that the distance between any two points or between any point and the opposite of another is never larger than the distance between s and v because there is no positive larger than v and no negative smaller than s. Therefore, the maximum value that any expression of the form |a+/-b| can have where a and b are points on the number line is the distance between s and v. D.</p>