1999 SOLUTIONS, PART I

Problem 1. Answer: 2019.

Problem 2. Answer: 2046.

Problem 3. Answer: 7.

Problem 4. Answer: 2.

Problem 5. Answer: 130.

Problem 6. Answer: 7.

Problem 7. Answer: bca. 210=1024, so 21999 is roughly 10600. 19992 is roughly 4·106. Observe
that 10=log21024 < log21999 < log22048=11. Hence the third number is between 1010 and 1011.

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: 5·62=180. Horse: 3·pi·52=75pi > 225. Goat: (1/2)22231/2/2,
which is between 180 and 225, since 1.5<31/2<1.8.

Problem 11. Answer: 7. The last digits of 777n 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! < 10G. Choose k so that 10G=gk=10100k, k=G/100. Thus 10G=gG/100 > gg > g!
On the other hand, 10G=10g+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/x2+2/x3=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: 1015< M < 1020. Let Mk 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 Mk+1=Mk+k(k+1)/2. So M is approximately 1018.

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(21/2)/(31/2). 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(31/2)/3=2/(31/2). The distance from the top vertex to
the bottom face is (4-4/3)1/2=2(21/2)/(31/2). The height of the bottom face is 1.