<p>EXrunner is right can u explain</p>
<p>(Cube problem, post#32)
If the original cube had each side = a, the volume shaved off is
2 (a^2)(0.5) + 2(a)(a-1)(0.5) + 2(a-1)^2 (0.5)</p>
<p>[Think of it as shaving the top & bottom faces first, then the left & right faces, then the remaining two faces].</p>
<p>Volume removed simplifies to 3a^2 -3a +1 = 169
Solve the quadratic equation for a, yields a=8. Check it : 8^3 - 7^3 = 169. So the remaining cube has each side=7, and its volume =7^3=343.</p>
<p>A different approach.
Blue towels are thrown in for the SAT flavor.
We can rephrase a question:</p>
<p>How many sets of towels selected from 10 numbered red towels are there, if a set can consist of any number of towels - from 0 (empty set; in old question form, all 10 out of 10 are blues) to 10 (old form: full red set - no blues).
Let's make a table with towels' numbers on top, and place 1’s in columns for the towels included in a set and 0’s for those that are not.</p>
<h2>1 2 3 4 5 6 7 8 9 10</h2>
<p>0 0 0 0 0 0 0 0 0 0 no red towels
0 0 0 0 0 0 0 0 0 1 set #1
0 0 0 0 0 0 0 0 1 0 set #2
0 0 0 0 0 0 0 0 1 1 set #3
0 0 0 0 0 0 0 1 0 0 set #4
0 0 0 0 0 0 0 1 0 1 set #5
0 0 0 0 0 0 0 1 1 1 set #6
So on …
Notice a pattern?
Well, not a fair question if you are not familiar with the binary system.
Briefly:
In a “regular” decimal system
247059 = 2<em>10^5 + 4</em>10^4 + 7<em>10^3 + 0</em>10^2 + 5<em>10^1 + 9</em>10^0
In binary system
111011 = 1<em>2^5 + 1</em>2^4 + 1<em>2^3 + 0</em>2^2 + 1<em>2^1 + 1</em>2^0.
Each combination of 0’s and 1’s in the table above represents a binary number, and they go in ascending order from 0 to – what do you think? – right, all the way to 1 1 1 1 1 1 1 1 1 1.
So we have as many towels sets as binary numbers in our table.
How do we count them? – by turning each number into its decimal equivalent:
0 => 0
1 => 1
10 => 2
11 => 3
100 => 4
and so forth.
Total number of sets is 1 more then the last number 1 1 1 1 1 1 1 1 1 1
(we start counting from 0).
To figure 1 1 1 1 1 1 1 1 1 1 out we could use a calculator or a formula for a partial sum of geometric progression. Do we really want to?
We need the next number: 1 1 1 1 1 1 1 1 1 1 + 1 = 1 0 0 0 0 0 0 0 0 0 0 = 2^10 in decimal.
That gives us the answer to the question: 2^10 = 1024.
++++++++++++++++++++++++++++++++++
Interesting connection with the binominal formula.
Remember what masta Xiggi said?
There are 45 ways to choose 2 different red towels out ot 10:
10C2 = 45.
We are looking for the number of all possible subsets from a set of 10 towels:
10C0 + 10C1 + 10C2 + … + 10C9 + 10C10.
Looks familiar – this buddies are all binominal coefficients!
10C0 + 10C1 + 10C2 + … + 10C9 + 10C10 = (1+1)^10
- binominal formula in reverse
(see <a href="http://talk.collegeconfidential.com/showthread.php?t=75251%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?t=75251</a> for more).
(1+1)^10 = 2^10 = 1024.</p>
<p>Is not math beautiful?!</p>
<p>Each side of original cube = a , its volume = a^3.
After shaving .5 off of each face, each sides gets shoorter .5 on each end.
New cube side = (a-1), its volume (a-1)^3.</p>
<p>a^3 - (a-1)^3 = 169
3a^2 - 3a - 168 = 0
a^2 - a - 56 = 0
a =-7 or a = 8.
I have hard time picturing a cube with a negative side (collapsed into itself? kinda like black cube?).</p>
<p>optimizerdad shaved his cube clean.</p>
<p>As helpful as it is, it does not stick to this cube.
Sorry.</p>
<h1>nmd</h1>
<p>Since all the answer choices are numeric, the answer should be the same for any values of p, r, s.
"greater then 2" is here just to make our life difficult - so we can't choose 2 (actually, we can - it does not change the answer).
Anyway,
p = 3, q = 5, r = 7
n = 3<em>5</em>7 = 105.
1 . 105
3 . 35
5 . 17</p>
<h1>7 . 15</h1>
<p>8 factors.
Kinda hands on approach.
Works often - if you see that the answers are numeric, in 99% of cases you can give any values to question's variables.
++++++++++++++++
Let's raise a bar.
Same question, but 7 prime factors:
n = p<em>q</em>r<em>s</em>t<em>u</em>v.
How many positive divisors now?</p>
<p>7c1 + 7c2 + 7c3 + 7c4 + 7c5 + 7c6 + 7c7 + 1 = 191 factors</p>
<p>nope, it ain't 191 ;<</p>
<p>What 'bout
n = p^1 * q^2 * r^3 * s^4 * t^5 * u^6 * v^7 ?</p>
<p>7c1 + 7c2 + 7c3 + 7c4 + 7c5 + 7c6 + 7c7 + 1 NOT= 191</p>
<p>Great approach though!
1 + 7c1 + 7c2 + 7c3 + 7c4 + 7c5 + 7c6 + 7c7 =
7c0 +7c1 + 7c2 + 7c3 + 7c4 + 7c5 + 7c6 + 7c7</p>
<p>Il bandito, you know the formula (a+b)^n.
Try to apply it to (1+1)^7 in algebraic form.
+++++++++++++++++++++
Challenge:
what's a faster and more generic solution (works for "primed up" too).</p>
<p>its not even geometry, its algebra.....
just assign variables x, y, z, (one of them being the angle we are looking for) and post equations. then use the identity technique to cancel the other two variables out, thus leaving you with the desired answer.</p>
<ol>
<li>A function is defined by f(x) = xe^(-2x) with domain (0,10)
a) find all x values where it is increasing and all values where it is decreasing ---> derivative is e^(-2x)-2xe^(2x)
b) give absolut min and max</li>
</ol>
<p>i guarantee that ^^ type of question is not on the SAT1...</p>
<p>Well actually, barely any of these except for the n = prs one would actually be on the SAT I. The towels wouldnt be and the triangle definitely wouldnt be.</p>
<p>There are 5 cubes placed inside one other, the sides of every smaller cube being 1/3 shorter then the sides of the preceding larger cube.
The side of the smallest cube is 3.14.
What's the ratio of the volumes of the largest and the smallest cubes?</p>
<p>cube small is 2/3 side as cube next
3.14 = 2/3 times x
x is 3/2 of 3.14 and corresponds to cube 2
cube three has a side of 3/2 of x or 3/2 of 3/2 of 3.14
cube four is (3/2)^3
cube five is (3/2)^4</p>
<p>so 81/16 is the ratio of the sides: or (81/16)/1 with 81/16 being cube 5
cube 5 side measure is 81x/16
volume of cube five is (81/16)^3 times 3.14^3
volume of original small cube is 3.14^3
ratio is (81/16)^3 which is 129.75 IS THIS RIGHT????</p>
<p>Sorry - it's not.</p>
<p>what is it???</p>
<p>You are right! (sorry, I misread).</p>