So we want the area inside the square outside the two circles. The overlap has area 2(400π/4 - 400/2), so requd area = 1600 - 400π + (200π - 400) = 1200 - 200π = 200(6 - π) = 200 x 2.858 = 571.6
| 1 400 | 9 161/3 |
| 2 502 | 10 572 |
| 3 164 | 11 945 |
| 4 62 | 12 792 |
| 5 660 | 13 |
| 6 156 | 14 |
| 7 320 | 15 |
| 8 819 |
1. Sum of such rationals < 1 is (1+7+11+13+17+19+23+29)/30 = 4. The rationals between 1 and 2 are all 1 larger, so sum 12. Similarly, those between 2 and 3 have sum 20, so total sum = 4 + 12 + 20 + ... + 76 = 400.
2. Each of the 9 non-zero digits can be in the number once or not. The order of the digits is then forced, so that gives 29 = 512 numbers, but the 1 zero-digit number and the 9 one-digit numbers are not allowed, giving 502 valid numbers.
3. Initially he won n out of 2n, then n+3 out of 2n+4 with n+3 > 0.503(2n+4), or 0.006n < 0.988, n < 164.6. Thus largest n is 164.
4. Suppose the first coefficient is n!/(r!(n-r)!), then 4/3 = (n-r)/(r+1), 5/4 = (n-r-1)/(r+2), so n = 62, r = 26.
5. We have 0.abcabcabc ... = abc/999. If abc is not divisible by 3 or 37, then this is in lowest terms. There are 333 multiples of 3, 27 multiples of 37, and 9 multiples of both, hence 999-333-27+9 = 648 which are neither. The 12 numbers which are multiples of 81 reduce to multiples of 3. There are no numbers which are multiples of 372, so we cannot get numerators which are multiples of 37. Thus 648 + 12 = 660 in total.
6. For c = 0, 1, 2, 3 or 4 and denoting c+1 by c', we see that 1abc + 1abc' has no carry iff a, b = 0, 1, 2, 3, 4 (total 125 possibilities). If c = 5, 6, 7, 8, then there must be a carry. Suppose c = 9. Again a, b = 0, 1, 2, 3, 4 have no carry (25 possibilities). If b = 9, then a = 0, 1, 2, 3, 4, 9 have no carry (6 possibilities). Grand total 125 + 25 + 6.
7. The perpendicular from D to BC must have length 16 (then area BCD = ½16·10 = 80). Hence the perpendicular from D to ABC has length 16 sin 30o = 8. Hence vol = 8·120/3 = 320.
8. Examining small terms we soon guess that an = (n-1)a2 - (n-2)a1 + ½(n-1)(n-2), or equivalently that an = ½n2 + an + b. But a19 = a92 = 0, so the polynomial must be (n-19)(n-92)/2 which has value 9·91 = 819 at n = 1.
9. Extend AD, BC to meet at X. Then B lies on the bisector of ∠AXB. But AP/PB = AX/BX. By similar triangles AX/BX = DX/CX = AD/BC = 7/5. Hence AP = (7/12)92 = 161/3.
10. Re(z/40) ∈ (0, 1), Im(z/40) ∈ (0, 1) simply means that if z = a + ib, then 0 < a, b < 40. Now 40/(a-ib) = 40a/(a2+b2) + 40bi/(a2+b2). So we require 0 < a, 0 < b, a2+b2 > 40a, or (a-20)2 + b2 > 202, and similarly a2 + (b-20)2 > 202.
So we want the area inside the square outside the two circles. The overlap has area 2(400π/4 - 400/2), so requd area = 1600 - 400π + (200π - 400) = 1200 - 200π = 200(6 - π) = 200 x 2.858 = 571.6
11. The effect of the two rotations is to change the angle of the line M from θ to θ - π/35 - π/27 = θ - 62π/945. In other words the two reflections give a rotation of 62π/945. Repeating n times gives a rotation of 62πn/945. We need this to be a multiple of π, so the smallest n is 945.
12. Take any path along cell edges from the top left to the bottom right (moving an edge at a time, always right or down). Each such path corresponds to a possible position. A path is a string of 5 moves down and 7 moves right, so 12C5 = 792 possible paths.
13.
14.
15. There are always enough powers of 2, we are constrained by powers of 5. We jump powers of 10, at multiples of 25. So 24! ends in 4 zeros (one 5 each from 5, 10, 15, 20), but 25! ends in 6 zeros. Similarly, multiples of 125 skip two, multiples of 625 skip three etc. So consider 7980! It has 1596 multiples of 5, including 319 multiples of 25, 63 multiples of 125, 12 multiples of 625, and 2 multiples of 3125. Hence total of 1596+319+63+12+2 = 1992. It has skipped 319+63+12+2 = 396.
© John Scholes
jscholes@kalva.demon.co.uk
6 Oct 2003
Last updated/corrected 6 Oct 03