| 1 300 | 9 33 |
| 2 137 | 10 120 |
| 3 182 | 11 486 |
| 4 480 | 12 19 |
| 5 588 | 13 1/930 |
| 6 193 | 14 373 |
| 7 70 | 15 462 |
| 8 112 |
1. 2·5·10·3.
2. Dist between centers = (142 + 182 + 212)1/2 = 31, so 19 + 31 + 87.
3. pn is not unless n = 3, pq is, p2q is not, p3q is not, pqr is not proper. So 6, 8, 10, 14, 15, 21, 22, 26, 27, 33.
4. Graph is quadrilateral with vertices (48, 0), (60,+-15), (80,0).
5. 3x2 + 1)(y2 - 10) = 507 = 3·132. The first factor is not divisible by 3, so the second must be. It cannot be 3, since 13 is not square, or 507 (nor is 517). So it must be 39, hence 3x2 = 12, y2 = 49.
6. Let distance of PQ from XY be h and XY = x. Then area XPQY = area WPQX, so (PQ + XY)h/2 = (PQ + WZ)(19-h)/2. Hence 2h = 19. Also AX + BY + DW + CZ = 2x - 38, so 2 AB = 4x - 38, or AB = 2x - 19. Hence area ABCD = (2x - 19)19. But 4 x area XPQY = (87 + x)19. Hence x = 106, so AB = 193.
7. 1000 = 2353, 2000 = 2453. So a = 2A5R, b = 2B5S, c = 2C5T, and max(A, B) = 3, max(A, C) = max(B, C) = 4, max(R, S) = max(R, T) = max(S, T) = 3. There are 10 ways of choosing R, S, T: 1 with all 3, 3 with one 2, 3 with one 1, 3 with one 0. C must be 4. Then there are 7 ways of choosing A, B: 1 with both 3, 3 with A = 3 and B not, 3 with B = 3 and A not. Thus 7 ways of choosing (A, B, C). Hence 7·10.
8. The condition is equivalent to 6n/7 < k < 7n/8. But (7n/8 - 6n/7) = n/56. So if n > 112, then there are certainly two integers in the interval (6n/7, 7n/8). For n = 112, we must have 96 < k < 98.
9. Put PC = x and use cosine rule: AC2 = 102 + x2 + 10x, AB2 = 102 + 62 + 10·6, BC2 = 62 + x2 + 6x. Hence 4x = 132.
10. Let T be time B takes to make 25 steps. Then B takes 3T to make 75, and A takes 2T to make 150. Suppose the escalator has N steps visible and moves n steps in time T. Then A covers N + 2n = 150, N - 3n = 75. Hence N = 120, n = 15.
11. Take numbers as N+1, N+2, ... , N+k. So k(2N+k+1) = 2·311. Since N >= 0, we have k < 2N+k+1, and hence k < (2·311)1/2. Also k divides 2·311. Largest such k is obviously 2·35.
12. Obviously m = n3 + 1, so (n + k)3 = n3 + 1, so 1 = 3n(nk + k2) + k3. If k <= 18, then rhs <= 54(18/1000 + 1/106) + 1/109 < 0.9721. But if n = 19, then putting k = 1/1000 gives 57(19/1000 + 1/106) + 1/109 > 1.
13. It is an easy induction to show that x20 gets to place 30 or later iff x20 iff it is the largest of x1, x2, ... , x30. Since it does not get to place 31 it must be less than x31. The chance that x31 is the largest of x1, ... , x31 is obviously 1/31. The chance that x30 is then the largest of x1, ... , x30 is 1/30.
14. We need the familiar a4 + 4b4 = (a2 + 2ab + b2)(a2 - 2ab + b2). Putting b = 3, we get a4 + 324 = ( (a-6)a + 18)( a(a+6) + 18). So most terms cancel leaving (58·64 + 18)/(-2·4 + 18) = 3730/10.
15. Letters represent areas. All triangles are similar, so P/Q = U/V = 441/440. Also P + U + 441 = Q + V + A + 440, so taking T = area of whole triangle, T = 441A. Thus sides of triangle area A are 1/21 x sides of whole triangle. So hypoteneuse of whole triangle is 4401/221. If its height from this side is h, then h - h/21 = 4401/2, so if two short sides have lengths a, b, we have ab = (4401/221)2/20 = 22·212, and a2 + b2 = 440·212. Hence (a + b)2 = 212(440 + 2·22) = 212222, and a + b = 21·22)
© John Scholes
jscholes@kalva.demon.co.uk
12 July 2003
Last updated/corrected 12 Jul 03