| 1 120, 719 | 9 615 |
| 2 101/200 | 10 83 |
| 3 484 | 11 (2 tan-12)/π |
| 4 12 | 12 7/9 |
| 5 154+40π/3 | 13 1155 |
| 6 12+12√2+4√3 | 14 127 (32/127) |
| 7 380 | 15 49/240 |
| 8 129 |
1. ((3!)!)! = 720!, so n = 719.
2. green/π = (22-12) + (42-32) + ... + (1002-992) = (2·1+1) + (2·3+1) + ... + (2·99+1) = 50 + 50·100. So ratio = 101/200.
3. sum = 7·34 + 6·21 + 5·13 + 4·8 + 3·5 + 2·3 + 1·2 = 484
4. sin x cos x = 1/10, so (sin x + cos x)2 = sin2x + 2 sin x cos x + cos2x = 1 + 1/5 = 6/5. Hence log(sin x + cos x) = ½ log 6/5 = ½ log 12/10 = (log 12 - 1)/2.
5. The box is 60. There are 4 quarter-cylinders length 3 and similarly for the other edges, total 3π + 4π + 5 π = 12π. There are 8 one-eighth spheres at the vertices, total 4π/3, and there are boxes height 1 on each face, 12+12+15+15+20+20 = 94. Total 154 + 40π/3
6. There are 24 half-faces, 24 with one edge a long-diagonal, and 8 equilateral (side √2). Hence 12 + 12√2 + 4√3.
7. Let M be midpoint of AC. D must lie on perpendicular at M. Suppose DM = x. Note that x does not have to be an integer. Put k = x2. k+36 = a2 and k+225 = b2 are squares. b-a = 1 gives 952-942. b-a = 3 gives 332-302. We must have b-a divides 189 = 337. b-a = 7 gives 172-102. b-a = 9 gives 152-62. b-a = 21 or more does not work. b-a = 9 also does not work because it gives D on AC. So perimeters 2·95+30, 2·33+30, 2·17+30 total 380.
8. a, a+d, a+2d, (a+2d)2/(a+d). So (a+2d)2-a(a+d) = 30(a+d) or 3ad + 4d2 = 30a + 30d. Hence d is multiple of 3. Also d(4d-30) = 3a(10-d). So 7.5 < d < 10. Hence d = 9. Hence a = 18. So 18, 27, 36, 48. Sum 129.
9. Suppose the 2-digit sum is d. We must have 1 ≤ d ≤ 18. For d ≤ 9, ther e are d+1 choices for the 2nd pair and one less for the 1st pair (no. cannot have leading 0). For d > 9, there are 19-d choices for each pair. Hence 1·2 + 2·3 + ... + 9·10 + 92 + 82 + ... + 12 = 615.
10. Take X on the line AM so that angle XBC = 7o. Since CA = CB, X must lie on the angular bisector of ∠C. So ∠BCX = 53o. Hence ∠XCM = 53o-∠ACM = 30o. Also ∠XMC = 7o+23o = 30o. Now XB is perpendicular to MC. But XM = XC, so it must be the perpendicular bisector of MC. Hence ∠CMB = ∠MCB = 83o.
11. If x is too small then cos2x > sin2x + sin x cos x. We have equality when x = k, where cos 2k = ½ sin 2k, or k = ½ tan-12. There is symmetry about π/4, so prob. of no triangle is k/(π/4).
12. The idea is shown in the diagram. Take BC' = BD and triangles ABC', CDB congruent. We have AB = 180 and AC' + AD = 280. So if M is the midpoint of C'D, then M = 140. Hence cos A = 140/180 = 7/9.
13. There are 2nCn binary strings length 2n with equal numbers of 0s and 1s. Half the others have more 1s. So (22n-2nCn)/2 with at least as many 1s. Hence (22n-2nCn)/2 strings length 2n+1 with more 1s. Half the strings length 2n-1 have more 1s, so 22n-2 strings length 2n have leading 1 and more 1s. Note that 2C1 = 2, 4C2 = 6, 6C3 = 20, 8C4 = 70, 10C5 = 252. So for strings with leading 1 we have the following numbers with more 1s:
length 1, 1
length 2, 1
length 3, 2+1=3
length 4, 4
length 5, 8+3=11
length 6, 16
length 7, 32+10=42
length 8, 64
length 9, 128+35=163
length 10, 256
length 11, 512+126=638
Total 1199.
Now 1984 = 11111000000, so all 44 numbers in the range 2004 to 2047 have more 1s than 0s. They must be excluded, giving 1155.
14. If n x 0. ... 251 ... = integer, then n x 0.251 ... = integer (where we simply delete the digits before 251. So it is sufficient to consider 0.251... . Obviously if n x 0.251 = integer, then n must be a multiple of 1000. So assume n < 1000. We require that [n x 0.251] = m, [n x 0.252] = m+1 for some m. Let {x} denote the fractional part of x. Then since n x 0.252 - n x 0.251 = n/1000, we must have {n x 0.251} > 1 - n/1000. If n is a multiple of 4, then {n x 0.251} = {n/1000}, so we need n > 500. If n = 1 mod 4, then {n x 0.251} = {0.25 + n/1000}, so n > 375. Similarly, if n = 2 mod 4, then n > 250, and if n = 3 mod 4, then n > 125. Thus the smallest candidate is 127, and we expect that 32/127 = 0.251... . That is easily checked.
15. Extend FD to meet the line BA at X. Then BX = BF. Now considering BXF, we have (AX/AB)(DF/DX)(CB/CF) = 1, so we get BF/BC = 720/867. Since BD is the angle bisector we have DC/DA = BC/BA. So (390+MD)/(390-MD) = 507/360. Hence MD/MC = 147/867. Finally, considering DFC, we have (ED/EF)(BF/BC)(MC/MD) = 1, so ED/EF = (867/720)(147/867) = 147/720 = 49/240.
© John Scholes
jscholes@kalva.demon.co.uk
1 Aug 2003
Last updated/corrected 15 Oct 03