*Problem 2.* Answer: 2046.

*Problem 3.* Answer: 7.

*Problem 4.* Answer: 2.

*Problem 5.* Answer: 130.

*Problem 6. *Answer: 7.

*Problem 7*. Answer: bca. 2^{10}=1024, so 2^{1999}
is roughly 10^{600}. 1999^{2} is roughly 4·10^{6}.
Observe

that 10=log_{2}1024 < log_{2}1999 < log_{2}2048=11.
Hence the third number is between 10^{10} and 10^{11}.

*Problem 8.* Answer: 6. Clearly 1,3,7,9,21,63 divide any such N.
These are the only divisors of 63.

*Problem 9.* Answer: 24. 20+17+11=48 counts each freshmen twice.

*Problem 10.* Answer: cow, goat, horse. Cow: 5·6^{2}=180.
Horse: 3·pi·5^{2}=75pi > 225. Goat: (1/2)22^{2}3^{1/2}/2,

which is between 180 and 225, since 1.5<3^{1/2}<1.8.

*Problem 11*. Answer: 7. The last digits of 777^{n} form
a periodic sequence: 7,9,3,1,7,9,3,1,...

*Problem 12.* Answer: 1 hour. A team of 2 Supermen, 2 Batmen and
2 Cinderellas will peel 3+4+5=12 buckets

in an hour. Therefore a team of 1 Superman, 1 Batman and 1 Cinderella
will peel 6 buckets in an hour.

Hence Superman will peel 6-5=1 bucket.

*Problem 13*. Answer: ACB. Suppose 100 calls are made, 10 of them
1 minute long, 10 - 5 min long,

30 - 10 min long, 30 - 20 min long, 20 - 30 min long. Plan A: 99·100+20·10·5=$109.

Plan B: 10+50+300+600+600=$156. Plan C: 0.8·156+25=$149.80.

*Problem 14.* Answer: 166. There are 6·5·4=120 words
with no repeating letters; 3·5·3=45 words in which one

letter appears twice (3 choices for the repeated letter, 5 choices
for the unrepeated letter, 3 choices for the

position of the unrepeated letter); 1 word EEE with a triple letter.

*Problem 15*. Answer: 3:04 pm. At 2 pm they are 2 miles away from
the lost paddle. D takes them back in 2/3 hour.

It takes them 2/5 hour more to get to the hat which is 2 miles down
the stream. The speed of the river is irrelevant.

*Problem 16*. Answer: 2/3.

*Problem 17.* Answer: 10G < g! < 10^{G}. Choose k
so that 10^{G}=g^{k}=10^{100k}, k=G/100. Thus 10^{G}=g^{G/100}
> g^{g} > g!

On the other hand, 10G=10^{g+1},g! (most factors are much bigger
than 10).

*Problem 18*. Answer: 29. Can pay: 6,7; 12,13,14; 18,19,20,21;
24,25,26,27,28; 30,31,32,33,34,35 and everything after that.

*Problem 19.* Answer: -4. 1-7/x+8/x^{2}+2/x^{3}=2(1/x-1/r)(1/x-1/s)(1/x-1/t).
So, 8=-2(1/r+1/s+1/t).

*Problem 20.* Answer: 1. (ii) implies that f(n) can be large; (iii)
implies that f(k)=k for k < f(n), and hence for all k.

*Problem 21.* Answer: 10^{15}< M < 10^{20}.
Let M_{k} be the number of steps required to order a list of k
numbers.

The (k+1)st number will not be reached by the computer before it orders
the first k numbers. The largest number of steps

to put the (k+1)st number in its place is k+(k-1)+...+1=k(k+1)/2. Hence
M_{k+1}=M_{k}+k(k+1)/2. So M is approximately 10^{18}.

*Problem 22.* Answer: yes,no,no. Assuming the usual checkered coloring,
one cannot remove squares of the same color.

*Problem 23.* Answer: 2bc cosw/(b+c). Let |AD|=x. The area of ABC
is (bc sin2w)/2. The areas of ADB and ADC

(which add up to the area of ABC) are (cx sinw)/2 and (bx sinw)/2.

*Problem 24.* Answer: 12. P(x)=k(x+1)x(x-1)(x-2) for some k. P(-2)=k(-1)(-2)(-3)(-4).
P(3)=k·4·3·2·1=P(-2)=12.

*Problem 25.* Answer: 1+2(2^{1/2})/(3^{1/2}). The
centers of the spheres form a regular tetrahedron with edge length 2. The

distance from a vertex of a face to the center of that face is 2(3^{1/2})/3=2/(3^{1/2}).
The distance from the top vertex to

the bottom face is (4-4/3)^{1/2}=2(2^{1/2})/(3^{1/2}).
The height of the bottom face is 1.