| 1 728 | 9 118 |
| 2 580 | 10 250 |
| 3 943 | 11 364/729 |
| 4 870 | 12 (336,8) |
| 5 763 | 13 160/3 sec |
| 6 495 | 14 √448 |
| 7 1/4 | 15 332/665 |
| 8 365 |
1. Suppose the first digit is even. Then there are 2 possibilities for the first digit, 4 for the last, 8 for the second and 7 for the third, total 448. Similarly, if the first digit is odd, then it must be 5 and there are 5·8·7 = 280 possibilities. Total 728.
2. Final x-coordinate = (12 - 32 + 52 - 72 + ... -392)/2 = -4(1 + 3 + 5 + ... + 19) = -400. Final y-coordinate = (22 - 42 + 62 - ... + 402)/2 = -2(3 + 7 + 11 + ... + 39) = -420. So distance = 20√(202 + 212) = 20·29 = 580.
3. N fish and M contestants. Then 6 fish average: 6(M - 21) = N - 19, and 5 fish average: 5(M - 8) = N - 108. Hence N = 943.
4. Put b = a + n, c = a + n + m. Then d = a + 2n + m. So 93 = bc - ad = n(n + m). So n = 1, n+m = 93 or n = 3, n+m = 31. There are 405 tuples (a, a+1, a+93, a+94) and 465 tuples (a, a+3, a+31, a+34)
5. p20(x) = p19(x-20) = p18(x-20-19) = ... = p0(x - (1 + 2 + ... + 20)) = p0(x-210). So coefficient is 3·2102 - 313·2·210 - 77 = 763.
6. n + (n+1) + (n+2) + ... + (n+8) = 9(n+4), n + (n+1) + ... + (n+9) = 5(2n+9), n + (n+1) + ... + (n+10) = 11(n+5). So N must be divisible by 9, 5 and 11. Conversely, 5·9·11 = 495 works.
7. Let the numbers be a1 > a2 > a3 > a4 > a5 > a6. There are 20 ways of choosing 3 of these for the dimensions of the brick, all equally likely. For the brick to fit, a1 must be a box side, and a6 must be a brick side. If a2 is a box side, then the brick certainly fits, giving 3 possibilities for the box (a1, a2, a3 or a1, a2, a4 or a1, a2, a5). If a2 is a brick side, then a3 must be a box side, giving 2 possibilities for the box (a1, a3, a4 or a1, a3, a5).
8. Let the subsets be A and B. For each element we have three choices (A, B or both). That gives each pair of subsets twice except for the case A = B = S. Hence (36 + 1)/2.
9. We must have n(n+1)/2 = 1993·1994/2 mod 2000, or (1993-n)(1994+n) = 0 mod 2000. Just one of 1993-n, 1994+n is odd, so either 1993-n or 1994+n is a multiple of 125 and the other is a multiple of 16. If 1993-n is a multiple of 125, then n = 118 mod 125 and 6 mod 16, and smallest is obviously 118. If 1993-n is a multiple of 16, then n = 9 mod 16 and 6 mod 125, so smallest is > 118.
10. V + 30 = E (Euler). VT + VP = 2E (counting edges). So V(T+P-2) = 60. Also, there are VT/3 triangles and VP/5 pentagons, so V(T/3 + P/5) = 32. Hence 32V(T+P-2) = 32·60 = 60V(T/3 + P/5). Hence 3T + 5P = 16. Hence P + T = 2. Hence V = 30.
11. Chance A wins the first game is ½(1 + ¼ + ¼2 + ... ) = 2/3. Let pn be chance A wins the nth game. Then it is an easy induction that pn = (1 + (-1)n+1/3n)/2
12. Working backwards we find that the requirement that Pn lies inside the triangle forces P6 = (28, 184), P5 = (56, 368), P4 = (112, 316), P3 = (224, 212), P2 = (448, 4), P1 = (336, 8).
13. We may assume A starts at (0, 0) and reaches X (t, 0) after t sec, and that A' starts at (0, 200) and reaches X' (3t, 200) after t sec. The center of the building O is at (50, 100). The line XX' is 100x - ty - 100t = 0, so O is a distance |100·50-100t-100t|/√(1002+t2) from the line. We want this to be 50, so 15t2 = 800t.
14. Let R be ABCD, center O. Let R' be PQRS, as shown. Let ∠ASP = θ. Then ∠BRS = ∠CQR = ∠DPQ = θ. So ASP and CQR are similar, but PS = QR, so AP = CR. Hence PR passes through O. But ∠PQR = 90o, so PR is a diameter of the circle PQRS. Similarly for QS. Hence O is the center of the circle PQRS.
We have AB + BC = (PS cos θ + RS sin θ) + (RS cos θ + QR sin θ), so perimeter R' = 28/(sin θ + cos θ) = 14√2/sin(θ+45o), which decreases as θ increases. Evidently the rectangle shown can be rotated, but the other rectangle with the same circle (and the same Q, S say) cannot. So the largest θ with a rotating rectangle occurs when P is the midpoint of AD. That implies area R' = ½ area R = 24, so PQ·QR = 24, PQ2 + QR2 = 64. Hence (PQ + QR)2 = 64+2·24 = 112.
15. We have AC = AU + CU = AT + CR = (AX - r) + (CX - r), where r is the inradius of ACX. Put AX = x, CX = z, AC = b. Then 2r = x + z - b. Similarly, if r' is the inradius of BCX and BX = y, then 2r' = y + z - a, so 2RS = |2r - 2r'| = |x - y - b + a| = |x - y - 1|. Now x2 = b2 - z2, y2 = a2 - z2, so x2 - y2 = b2 - a2, and x - y = (b2 - a2)/c = 3987/1995. Hence RS = 996/1995.
© John Scholes
jscholes@kalva.demon.co.uk
3 Oct 2003
Last updated/corrected 3 Oct 03