| 1 8 | 9 (1,5,1),(100,20,100) |
| 2 21 | 10 75/98 |
| 3 667 | 11 248 |
| 4 260 | 12 177 |
| 5 26 | 13 700/31 |
| 6 997 | 14 4/7 |
| 7 1/4 | 15 927 |
| 8 12 - 3·37(1/3) |
1. We need 2n or 5n to have a 0. It is easy to check the first is 58 = 390625.
2. BCDE is a rectangle sides 2u, 2v, so area 4uv. The triangle ABE has base BC length 2u and height u-v, so area u2 - uv. Hence u(u+3v) = 451 = 11·41. Hence u = 11, v = 10.
3. We must have (1998/3)(m/n) = 1, or m/n = 1/666. Hence m = 1, n = 666.
4. Let the sides of the squares be a, b, ... , i (in ascending order of size - a is not marked in the diagram). We have a + b = c (1), a + c = d (2), c + d = e (3), d + e = f (4), b + c + e = g (5), b + g = h (6), a + d + f = i (7), f + i = g + h (8). Using (1) - (7) we can solve in terms of a, b, then (8) gives 5a = 2b, so a = 2, b = 5 (must have no common factor). Hence f = 25, h = 33, i = 36, sides are 61, 69.
5. Suppose one box has n marbles, b of which are black, and the other 25 - n marbles, B of which are black. So bB/(n(25-n)) = 27/50. n(25-n) must be a multiple of 5, so n must be a multiple of 5. Hence n = 5, 10, 15 or 20. Without loss of generality, n = 5 or 10 (just swap boxes). If n = 5, then bB = 54. So b divides 54 and <= 5, so b = 1, 2, or 3 giving B = 54, 27, 18. Thus b = 3, B = 18, and prob both white = (2/5)(1/10) = 1/25. If n = 10, then bB = 81, so b = B = 9. Hence prob both white = (1/10)(2/5) = 1/25.
6. We have (m+n)/2 = √(mn) + 2, so mn = m2/4 + n2/4 + mn/2 - 2m - 2n + 4. Hence (m+n) = 2 + 2( (m-n)/4 )2. So m-n must be a multiple of 4. 4 is too small, it gives m = 4, n = 0. 999 is too big, it gives m = 1000000. But the 997 values 4·2, 4·3, ... , 4·998 all work.
7. 1/z = 5-x, y = 29 - 1/x. Hence x(29 - 1/x) = (5 - x), so x = 1/5, y = 24, z = 5/24 and z + 1/y = 1/4.
8. When point down the liquid dimensions are 3/4 those of the container, so its volume is 27/64 that of the container. Hence the empty space has volume 37/64 that of the container. When point up, the empty space occupies a cone, whose height must be (37/64)1/3 that of the container. Hence depth of liquid is 12(1 - (37/64)1/3).
9. Since log(2000xy) = 3 + log(2xy), the first two equations give log y = 1 or log x = log z, ie y = 10 or x = z. But y = 10, does not satisfy the first equation, so x = z. The last equation gives 2 log x - log x log x = 0, so x = 1 or 100. If x = z = 1, then the second equation gives y = 5. If x = z = 100, the second equation gives y = 20. So the solutions are (1, 5, 1) and (100, 20, 100).
10. Let s = x1 + x2 + ... + x100. Then 2xk = s - k. Summing, s = 50s - (1 + 2 + ... + 100)/2, so s = 25·101/49.
11. We have (1+2+4+8)(1+ 1/5 + 1/25 + 1/125) + (1+5+25+125)(1 + 1/2 + 1/4 + 1/8) -1 + (1/10 + 1/20 + 1/40 + 1/50 + 1/100 + 1/200 + 1/250 + 1/500 + 1/1000) + (10/1 + 20/1 + 40/1 + 50/1 + 100/1 + 200/1 + 250/1 + 500/1 + 1000/1) = 2480.427.
12. From the first two, f(x) = f(1760+x). From the second two, f(x) = f(1056+x). gcd(1056, 1760) = 352, so f(x) = f(x+352). Then f(x) = f(398-x) gives f(x) = f(46-x). Conversely, it is easy to check that f(x) = f(46-x) and f(x) = f(352+x) imply the relations given. Evidently, f(x) = f(352+x) means the values repeat after any block of 352. Take that block centered on x = 23. We can choose separately the values at x = 23-176, 23-175, ... , 23-1, 23. Then the reflection determines the values at 23+1, 23+2, ... , 23+176. The period then determines all other values. So we get at most 177 different values. If we take them all different, they certainly appear in x = 0, 1, ... , 999, which contains two complete periods.
13. The fire truck can drive along the axis and then head across the prairie in a straight line. If it starts straight across the prairie it can reach anywhere in a circle radius 1.4. If it drives 5 along the road, the circle has radius 0. For points in between the circle is scaled proportionately. So the envelope is the four-pointed star. The part in the first quadrant is the difference between an isosceles right-angled triangle hypoteneuse 5√2 and an obtuse-angled triangle with the same base. The height of the second triangle is 5/√2 tan(45o - x), where tan x = 7/24 (the right-angled triangle has sides 5, 1.4 and hence 4.8 - it is a 7, 24, 25 triangle). So tan(45o - x) = 17/31. Hence total area = 700/31.
14. AQ = 2 cos x, so AB = (1 + 2 cos x), so (1 + 2 cos x) sin x/2 = 1/2. But sin 3x/2 - sin x/2 = 2 cos x sin x/2, so (1 + 2 cos x) sin x/2 = sin 3x/2. Hence 3x/2 = 30o, so x = 20o. Hence angle APQ = 140o, and angle ACB = 80o, so ratio = 4/7.
15. Work backwards. It is easy to see that shortly before the end the pile is 1998, 2000, 1999. Move 1999 to the top and put 1997 above it, giving 1997, 1999, 1998, 2000. Move 2000 to the top and put 1996 above it, giving 1996, 2000, 1997, 1999, 1998. Move 1998 to the top and put 1995 above it, giving 1995, 1998, 1996, 2000, 1997, 1999. Now denote the state by (n, m), where the pile has n cards and there are m below the 1999. We see that (n, m) is preceded by (n+1, m-1) for m > 0, and (n, 0) is preceded by (n+1, n-1). Thus from (3, 0) we went to (4, 2), (5, 1), (6, 0), (7, 5), ... , (12, 0), ... , (24, 0), ... , (48, 0), ... , (96, 0), ... , (192, 0), ..., (384, 0), ... , (768, 0), ... , (1536, 0), (1537, 1535), (1538, 1534), (1539, 1533), ... , (2000, 1072). Hence at the start there are 2000 - 1073 = 927 cards above 1999.
(C) John Scholes
jscholes@kalva.demon.co.uk
4 Aug 2003
Last updated/corrected 4 Aug 03