| 1 81649 | 9 417/512 |
| 2 201, 499 | 10 784 |
| 3 898 | 11 98/243 |
| 4 60/7 | 12 69/32 |
| 5 253 | 13 64/5 |
| 6 25 | 14 15,75,105,165 and 360-these |
| 7 √7250 | 15 372+39√6 |
| 8 429 |
1. The two digit squares are 16, 25, 36, 49, 64, 81. So if 16 occurs in the number, then it must be followed by 4, which must be followed by 9 and nothing can follow 9. Similarly, 16 can only be preceded by 8, which cannot have predecessor. 64 can also be preceded by 36, which cannot have a predecessor. 25 cannot have successors or predecessors. So the only numbers that cannot be extended at either end are 25, 3649, 81649. The longest is obviously 81649.
2. 601 to 800 study French, 1601 to 1700 study Spanish. So smallest for both is 201 with 400 just French, 1400 just Spanish. Largest is 499, with 301 just French, 1201 just Spanish.
3. We find a5 = 267, a6 = -a1, a7 = -a2, a8 = -a3, a9 = -a4, a10 = -a5, a11 = a1, a12 = a2, a13 = a3, a14 = a4 etc. So a531 = a1, a753 = a3, a975 = a5 and a1 + a3 + a5 = 211 + 420 + 267 = 898
4. The line through (8,6) parallel to OQ is 10y = 3x + 36, and this meets OP at (16/7, 30/7). But that must be the midpoint of OP. So P is (32/7, 60/7). The distance to (8,6) is √((24/7)2 + (18/7)2) = 30/7. Hence PQ = 60/7.
5. Clearly (4, 5, 9, 14, 23, 37, 60, 97, 157, 254) does not have the triangle property (each element from 9 on is the sum of the previous two). But if a1 < a2 < ... < a10 does not have the property, then a10 ≥ a9 + a8 ≥ 2a8 + a7 ≥ 3a7 + 2a6 ≥ ... ≥ 34a2 + 21a1 ≥ 34·5 + 21·4 = 254.
6. Let the large square have side 2, the small square side x. Then the radius is √2. The lines containing vertical sides of the small square pass a distance x from the center. So by Pythagoras, 2 = (1 + 2x)2 + x2, so (5x-1)(x+1) = 0, so x = 1/5. So ratio = 1/x2 = 25.
7. Let B be the vertex with angle 90o. Suppose it is a distance x from the points of contact of the incircle. Then chasing around the triangle using the fact that the tangents from a point have equal length, we get x = (90 + 120 - 150)/2 = 30. Evidently this is also the radius of the incircle. The other two triangles are similar. The top one has sides 30, 40, 50 and inradius 10, and the right-hand one has sides 45, 60, 75 and inradius 15. We can regard the dotted line as the hypoteneuse of a right-angled triangle with vertical side 60+10-15 = 55, horizontal side 60+15-10 = 65. Hence length √(552 + 652) = √7250 = 5√290.
8. The graph of f for x ∈ [1,3] is piecewise linear with f(1) = f(3) = 0 and a peak at f(2) = 1. The graph for x ∈ [3,9] is a 3x larger copy with peak at 3, the graph for x ∈ [9,27] is 3x larger again, with peak at 9 and so on. f(x) is linear from f(2·729) = 729 to f(2001) = 729-543 = 186. f first reaches this value between x = 243 and x = 486. f(243) = 0 and f(486) = 243, so f(x) = 186 at x = 243 + 186 = 429.
9. We use the inclusion/exclusion principle. There are 4 possible red squares as shown. Let pij...k be the probability of getting i and j and ... and k. Obviously p1 = p2 = p3 = p4 = 1/16, so ∑ pi = 1/4. Similarly p12 = p13 = p24 = p34 = 1/64, p14 = p23 = 1/128, so ∑ pij = 5/64. We have pijk = 1/256, so ∑ pijk = 1/64. Finally p1234 = 1/512. Hence p(none) = 1 - 1/4 + 5/64 - 1/64 _ 1/512 = 417/512.
10. (106 - 1) = (103 + 1)(103 - 1), so 106 - 1 and hence also 106n - 1 is certainly divisible by 1001. But 7 divides 1001 and 10m - 1 is not divisible by 7 for m not a multiple of 6. Obviously 10m is never divisible by 7. Thus 10m(10n - 1) is divisible by 1001 iff n is a multiple of 6. So for j = 0, there are 16 values of i (6, 12, ... , 96). Similarly for j = 1, 2, 3. For the next 6 values of j there are 15 values of i, total 6·15. For the next 6, there are 14 each, total 6·14, and so on up to the 6 values 88 to 93 for which there is one each, total 6·1. For j > 93, there are no values of i. Hence total = 64 + 6(15 + 14 + ... + 1) = 64 + 720 = 784.
11. By symmetry the prob of more wins than losses equals the prob of more losses than wins. We calculate the prob of the same number of wins and losses. Prob of no wins, no losses = 1/36. Prob of 1 win, 1 loss = 6·5/36. Prob of 2 wins, 2 losses = (6C2)(4C2)/36 = 15·6/36. Prob of 3 wins, 3 losses = (6C3)/36 = 20/36. So prob same no. = (1+ 30 + 90 + 20)/729 = 141/729 = 47/243. Hence prob more wins than losses = prob more losses than wins = 98/243.
12. The side length of each small tetrahedron on the face of the original tetrahedron is 1/2 the side length of the original, so its volume is 1/8. There are 4 of them so total volume 1/2. The new figure has 24 faces. The second time, each small tetrahedron has volume 1/64, so their total volume is 24/64 = 3/8. The new figure has 3·24 = 72 faces which are subsets of faces of the previous figure and another 72 outside, total 144. The tiny tetrahedra have vol 1/512, so total 144/512 = 9/32. Hence total vol = 1 + 1/2 + 3/8 + 9/32 = (32 + 16 + 12 + 9)/32 = 69/32.
13. Extend AB, CD to meet at E. Then triangles ECB, EBD are similar. Hence BD/ED = CB/EB, so 10/ED = 6/(ED-8). Hence ED = 20. Also EC/CB = EB/BD, so EC = 7.2. Hence CD = 12.8 = 64/5.
14. We have sin 28θ - sin 8θ = 0. Hence 28θ = 8θ + 360n or 28θ + 8θ = 180 + 360n. In the first case we have cos 28θ = cos 8θ. But cos 28θ - cos 8θ = 1 (*). So we must have 36θ = 180 + 360n. Hence θ = 5, 15, 25, ... , or 355 (or 5 + 10n).
(*) gives - 2 sin((28θ+8θ)/2) sin((28θ-8θ)/2) = 1 or sin 18θ sin 10θ = -1/2. Putting θ = 5 + 10n, leads to sin(100n+50) = (-1)n+1/2. Now n odd gives θ = 15 or 75 mod 180 and n even gives 105 or 165 mod 180.
15. This is messy. Take coordinates with vertices of the cube at (0,0,0), (8,8,8), (8,0,0), (0,8,0), (0,0,8) etc. Bore the hole from (0,0,0) to (8,8,8). The long edges of the hole are from A (6,8,8) to B (0,2,2), from C (8,6,8) to D (2,0,2), and from (8,8,6) to (2,2,0). The plane ACDB has equation 2z = 2+x+y. So it meets an edge of the cube at E (0,0,1). ACDEB is one face of the tunnel. We find AC = 2√2, AB = 6√3, BE = √5. After some calculation, area ACDEB = 13√6. So the tunnel faces have total area 39√6.
There are 3 faces which are just original cube faces less a small triangle. These each have area 62. Another three are original cube faces less a small kite shape removed from the corner. These also have area 62. So total excluding tunnel faces = 372.
© John Scholes
jscholes@kalva.demon.co.uk
10 October 2003
Last updated/corrected 23 Mar 04