*Problem 2.* Dividing all three by (9!)^{ 1/2 }
yield 10, 9· 10^{ 1/2}, and 10^{ 1/2 }·
11^{ 1/2 }, respectively. Answer: e.

*Problem 3.* When x<=-2, |x|=-x and |x+2|=-x-2, sp
the equation holds. It fails in all other cases. Answer: a.

*Problem 4.* Since log(2)+log(3)=log(6), the expression is
equal to 10^{ log(6)}=6. Answer: b.

*Problem 5.* 43^{ 2}=1849 is the only perfect square in
the 1800's, so he was born in 1849-43=1806. Answer: a.

*Problem 6. * Every blab is a blib, and some of these are blubs,
so (since there are some blabs) some blibs are blubs. Answer: c.

*Problem 7*. If D=# of dogs and C=# of cats, then
D+24=2C and D=2(C-x). Thus, D+24=D+2x, so x=12. Answer: b.

*Problem 8.* 29=(27+36+25+28)/4. Note that
29 is also the mean of all 5 numbers. Answer: d.

*Problem 9.* Label the trapezoid ABCD, with AB of length 20 on
the bottom and CD of length 10 on the top. To find
the length of the diagonal AD, drop a perpendicular from
D onto AB, and let X be the place where it intersects.
Then DXB is a right triangle, so DX ^{ 2}=89-5^{ 2}.
Thus, DX=8. Since DXA is a right triangle, AD^{ 2}=15^{ 2}+
8^{ 2}. Answer: c.

*Problem 10.* We count the cases when Romeo (R) is to the left
of Juliet (J) and Caesar (C) is to the left of Brutus (B) and multiply by 4.
Label the positions 1,2,3,4,5,6.
If R is in position 1, then J is in 2. If in addition, C is in 3 then there
are 4 possibilities; if C is in 4, then there are two possibilites
and nothing else is possible. So there are 6 possibilities when R is in 1.
When R is in 2, J is in 3. If C is in 1, then there are 6 possibilities,
but if C is in 4, there are 2 cases, giving a total of 8 cases when R is in 2.
When R is in 3, J is in 4. If C is in 1, then there are 4 cases, while if
C is in 2, then there are 4 cases, giving a total of 8 cases.
When R is in 4, J is in 5. If C is in 1, then B can be in two places, giving
a total of 4 cases. If C is in 2, then B must be in 6, giving 2 cases,
and if C is in 3, then B must also be in 6, giving 2 cases
for 8 cases total when R is in 4. When R is in 5, J is in 6.
If C is in 1, there are 4 cases; if C is in 2, there are 2 cases, giving
6 cases. Thus, there are 6+8+8+8+6=36 cases satisfying
these addtional constraints, so there are
4· 36=144 cases altogether. Answer: d.

*Problem 11*.
Multiplying by sin(x) and simplifying yields sin^{2}(x)=cos(x).
Let u=cos(x). Then, since sin^{2}(x)+cos^{2}(x)=1,
u=1-u^{2}. Applying the quadratic formula and throwing
out a root since |u|<=1 yields u=(5^{ 1/2}-1)/2. Answer: c.

*Problem 12.*
The total amount of chocoalate in a bunny is proportional to the CUBE
of its height. So, assuming that the amount of chocolate in a 1-inch bunny is k,
the amount of chocolate consumed by
F is (5k)^{ 3}=125k^{3} , by M is 10· (2.5k)^{3}, and by
B is 100· k^{3}. Clearly, B< F and F< M since 2.5^{3}>12.5.
Answer: c.

*Problem 13*. The proportion 17:27 implies that we want 510 lbs of Copper
and 810 lbs of Tin. If A and B represent the amounts of each alloy, we have
two equations in two unknowns: 1/3 A+2/5B=510 and 2/3 A+3/5B=810.
Thus A=270 and B=1050. Answer: e.

*Problem 14.*
If S is the sum of the integers from k+1 to k+n, then
S+100 is the sum of k+n+1 to k+2n. Each of the n terms in the latter
sum is n more than the corresponding entry in the former sum.
That is, n^{ 2}=100, so n=10.
Answer: a.

*Problem 15*. Larry (L) and Curly (C) drive for 3 hours (75 miles). Then C got out,
and L drives backward for 2 hours. L picks up Moe and they drive for 3 more hours.
Total time=8 hours. Answer: d.

*Problem 16*. Suppose that the radius of the circle is r.
Let s be the side length of the square inside the semicircle,
and let t be the side length of the square inside the full circle. We
want to calculate s^{2}/t^{2}. In the semicircle, there is
a right triangle with sides s,s/2, and hypotenuse r. Thus, r^{2}=5s^{2}/4,
while the full circle gives a right triangle with sides t/2,t/2 and hypotenuse r.
So r^{2}=t^{2}/2. So
s^{2}/t^{2}=2/5.
Answer: c.

*Problem 17.* (0.1 mm/day)(1 km/10^{6} mm)(1 mile/1.6 km)(1 day/24 hours)=
2.6x10^{-9} mph. Answer: e.

*Problem 18*. They start fighting at 1:00pm. At that time they have
completed 1/3+1/4=7/12 of the job. Tom starts painting again at 1:10
and finishes in (5/12)· 3=5/4 hours. Answer: b.

*Problem 19.* Let t be the side length of the triangle.
There is a right triangle with sides h,t/2 and hypotenuse t. So
h^{2}=3t^{2}/4. Thus, the area of the triangle
is 1/2 th=h^{2}/3^{1/2}. The area of the square is s^{2},
so h/s=3^{1/4}. Answer: a.

*Problem 20.*
(4.5)^{2}+(1.5)^{2}=5/8(6^{2}). Answer: c.

*Problem 21.* Since 3^{8} > 5000, the only numbers are
1^{8}+ 1^{8},
1^{8}+ 2^{8},
1^{8}+ 3^{8},
2^{8}+ 2^{8},
2^{8}+ 3^{8}. Answer: a.

*Problem 22.* 2008! is the smallest factorial that is divisible by
2^{2001}. There are 1004 numbers <= 2008 that are divisible by 2. Of these,
502 are divisible by 4. Of these, 251 are divisible by 8. Of these 125 are divisible by 16.
Of these 62 are divisible by 32. Of these, 31 are divisible by 64.
Of these, 15 are divisible by 128. Of these,
7 are divisible by 256. Of these, 3 are divisible by 512 and finally 1 is divisible by 1024.
Answer: d.

*Problem 23.* Any root of the polynomial will divide 420=1· 4·
3· 5· 7. Thus, the largest number of distinct roots occurs
with the polynomial (x-1)^{3}(x-2)^{2}(x-3)(x-5)(x-7)
or
(x-1)^{4}(x-4)(x-3)(x-5)(x-7). In either of the two cases,
there are 5 distinct roots. Answer: b.

*Problem 24.*
Huey can get any of 4 duck's hats, and each of these are symmetric.
Assume that Huey gets Dewey's hat. If in addition Dewey gets H's hat,
then there 2 possibilities for which hat Donald gets. If Dewey does NOT
get H's hat, then there are 3 symmetric possibilities depending on whose
hat he gets. Assume Dewey gets Donald's hat. Then there are 3 possibilities.
Thus, by symmetry, there are 2+3· 3=11 arrangements that start
with Huey getting Dewey's hat. By symmetry again, there are 4· 11=44
possibilities in all. Answer: c.

*Problem 25.*
Let r be the distance from the center of the circle to a vertex and let s be
the length of one side of the original octagon. Let T denote the isocoles triangle
with two sides of length r and having base s. The area of the whole octagon
is 8 times the area of this triangle.
By the Law of Cosines, s^{2}=
r^{2}+
r^{2}-
2r^{2}cos(360°/8)=r^{2}(1- 2^{1/2}).
Now let A be the midpoint of one side of the
octagon, let B be the midpoint of an adjacent side, and let
C be the center of the circle. Let D denote the vertex of the large
octagon connecting the adjacent sides. Let x
denote the length of BC. The area of the smaller octagon is 8 times
the area of triangle ABC, so we are asked to find the ratio of the area of ABC
to the area of T. Since the triangle ABC is similar to T, this ratio is
equal to (x/r)^{2}. But CBD is a right triangle with sides x, s/2 and
hypotenuse r. So r^{2}=x^{2}+s^{2}/4.
Combining this with the equation above gives
x^{2}=r^{2}(2-2^{1/2})/4. Answer: d.