<p>1) S is the sum of the 1st 100 positive EVEN integers, and T is the sum of the first 100 consecutive positive integers. S is what percent greater than T?</p>
<p>2) If kn does not equal k, and n=1/k, which of the following expressions is equivalent to (1-k)/(1-n)?
A)-n
B)-K
C)1
D)k
E)n</p>
<p>3) The 1st 2 numbers of a sequence are 1 and 3 and the third number is 4. In general, every # after the second is the sum of teh two # immediately preceding it. How many of the first 1000 #'s in this sequence are odd?</p>
<p>4) Circle C has radius of squareroot of 2. Squares w/ sides of length ` are to be drawn so that, for each square, one vertex is on circle C and the rese of the squre is inside circle C. What is the greatest number of such squares that can be drawn if the squares do not have overlapping areas?</p>
<p>A)NONE
B)1
C)2
D)3
E)4</p>
<p>5) If the ratio of 2 positive integers is 3 to 2, which of the following statments about these integers cannot be true?</p>
<p>A) their sum is an odd integer
B)there sum is an even integer
C)there product is divisble by 6
D)their product is an even integer
E)there product is an odd integer</p>
<p>Your math teacher should have gone over how Gauss discovered these formulas as a grade school student.</p>
<p>If your teacher didnt tell you the little Gauss story just look at the symmetry in the sequence and it should be easy to see how he deduced it: 2+12 = 4+10 = 6+8 = 14</p>
<p>I will leave the other 4 to other contributors.</p>
<p>Good call for 1.
For 2:
It's faster to solve algebraically. n = 1/k, so plug in for the expression 1-k/1-n. You get 1-k / 1-(1/k). Multiply that by k/k, and you get: k(1-k)/(k-1). factor out -1 from the numerator and you get: -k (k-1) / (k-1). The k-1 cancels out, giving -k.</p>
<p>5) E)there product is an odd integer
Heres why: pick #s
for A: (3,2) 3+2 = 5 odd
for B: (30,20) 30+20 = 50 even
for C and D: (3,2) 3*2 = 6 6 is divisble by 6 and is even.</p>
<p>Infact any product that is divisble by 6 will be even and therefore NOT odd, making e the obvious choice.</p>
<p>so, if we do the first few... 1, 3, 4, 7, 11, 18...
they go odd odd even odd odd even, recursively. so 2/3 are odd, and they are the first two,
so for 999 terms, 666 are odd and so is the thousandth term, making 667.</p>
<p>If the ratio is 3:2, then no matter what the numbers, its basic "unit" is going to be 3:2. I suck at explaining. But anyway, I would think:</p>
<p>3+2 = 5 = odd number, so A is out.
2(3+2) = 10 = even number, so B is out. (6:4 = 3:2)
3x2 = 6 = divisible by 6, so C is out.
3x2 = 6 = even integer, so D is out.
If the basic 3:2 unit has an even product, then any two numbers that have a 3:2 ration will necessarily be a multiple of 6, and will therefore be even. So the answer is E, because the product can never be odd.</p>
<p>Thats the kind of thing yale wants to hear jimbob ;>
I don't think #4 is doable with the given information, or i just don't understand the question</p>
<p>4) I'll assume that you hit the wrong key and that you meant that each side =1. </p>
<p>Draw a circle, and then draw a horizontal and a vertical line to represent the radii. (You would have a circle divided into four parts by a t.) Remember that triangle ratio for 45-45-90 triangles? The 1-1-radical 2. You put two of those together to make a square, right? So just place the unit squares so that their diagonal lines up with a radius. That way, you can fit exactly FOUR squares.</p>
