| 1 869 | 9 144 |
| 2 968 | 10 994 |
| 3 750 | 11 947 |
| 4 675 | 12 137 |
| 5 40/243 | 13 905 |
| 6 160m | 14 490 |
| 7 925 | 15 108 |
| 8 334 |
1. (n-1)n(n+1)(n+2) + 1 = ( n(n+1) - 1)2
2. There is just one convex polygon for each subset of n points for n = 3, 4, ... , 10. So no. = total no. of subsets - no. of subsets size 0, 1, 2 = 1024 - 1 - 10 - 45.
3. Decimal = (100d + 25)/999, so 37n = 3000d + 750. So certainly 750 divides n. If n = 750m, then 37m = 4d + 1. That has unique solution d = 9, m = 1, for d a digit.
4. Middle integer n, then 3n is square, so n = 3k2 for some k. Also 5n is cube, so 15k2 is cube, so k is a multiple of 15. Smallest value 15 gives n = 3·152.
5. 5p(1-p)4 = 10p2(1-p)3, so 1-p = 2p, and p = 1/3. Prob of 3H is 10p3(1-p)2.
6. Applying cosine rule to velocity triangle, we get that D's vel relative to C is k where 72 = 82 + k2 - 8k. So k = 3 or 5. Hence largest possible k is 5. Time 100/5 = 20s.
7. Let the ap be a-d, a, a+d, squares a2-2ad+d2, a2, a2+2ad+d2. So d(2a - d) = 264, d(2a + d) = 296. Hence d = 4, a = 35, so 1225 = 300 + k.
8. Label equations given (1), (2), (3). Then (1) - 2 (2) + (3) = 2(x1 + ... + x7), (3) - (2) = 5x1 + 7x2 + 9x3 + ... . Hence 7x1 + 9x2 + 11x3 + ... = (1) - 3 (2) + 2 (3) and required value = (1) - 3 (2) + 3(3).
9. The last digits of n and n5 are the same. Hence last digit of lhs is same as that of 3 + 0 + 4 + 7 = 4. Hence last digit of k is 4. Also 133 = 1 mod 3, 110 = -1 mod 3, 84 = 0 mod 3, 27 = 0 mod 3, so lhs = 0 mod 3. Obviously k > 133. So smallest possibility is 144, next is 174. Now 115 = (10 + 1)5 = 105 + 5·104 + 10·103 + 10·102 + 5·10 + 1 = 161051, so 1105 < 2 1010. Obviously, 27 and 84 < 100, so 275 and 845 < 1010. Similarly, 1335 < (1331/10)5 = 1115/105 < 5 1010. Hence lhs < 1011. But 1702 = 28900 > 28000, 1704 > 780000000 > 7 108 and 1705 > 1011. So only possibility is 144.
10. wlog c = 1. Obviously a, b must be about 31.5 (remember |a - b| < 1). Suppose angle B = 90o. Then a2 = b2 + c2, so a = 9951/2, b = 9941/2. Hence cot B = 0, cot C = 1/cot A = 9941/2. Hence expr = 994. So if cot C/(cot A + cot B) is independent of a, then it must have value 994. Proving that it is independent is harder, but not strictly necessary.
11. Optimum is evidently n+1 terms = 1, n terms = 1000, 999, 998, ... (as far as necessary). n = 1 gives 2 of 1, 1 each of 881, ... , 1000, mean 924.9, diff 923. n = 2 gives 2 each of 942, ... , 1000, mean 946.9, diff 945. n = 3 gives 3 each of 962, ... , 1000, mean 948.6, diff 947. n = 4 gives 4 each of 972, ... , 1000, mean 945.3, diff 944. In the general case the mean is (1001 - (120-n)/n + 1000) (120 - n)/2 + n+1)/121, so we wish to maximise (2002 - 120/n)(120 - n)/2 + n and hence to maximise -1000n - 7200/n, or equivalently to minimise 36/n + 5n. Easily seen this is at n = 3.
12. Using the cosine formula twice we find that if a triangle has sides a, b, c and median length m to the middle of the side length a, then 4m2 = 2b2 + 2c2 - a2. So the square of the median from vertex 7, 18, 41 to the side 13 is 577/4 and the square of the median from the vertex 27, 36, 41 to the side 13 is 3881/4. Hence 4 x requd square = 577/2 + 3881/4 - 412.
13. We show first that we can choose at most 5 numbers from {1, 2, ... , 11} such that no two have difference 4 or 7. Translating if necessary, we make take the smallest number to be 1. That rules out 5,8. Now we can take at most one from each of the pairs: 2,9; 3,7; 4,11; 6,10. 1989 = 180·11+9, so that means we can pick at most 5·181 = 905. If we pick 1, 3, 4, 6, 9, then that works and moreover it allows us also to pick 11+1, 11+3, 11+4, 11+6, 11+9. So 905 is also possible. Thanks to Bugz Podder for this.
14. We have (i - 3)3 = -18 + 26i, (i - 3)2 = 8 - 6i. The imaginary part must vanish, so 26a3 - 6a2 + a1 = 0. If a3 = 1, this implies a2 = 5, a1 = 4; if a3 = 2, it implies a1 = 9, a1 = 2; and we cannot have a3 > 2. The 4-digit number 1540 in this base = 10, and 2920 in this base = 30. So required sum is 10 + 11 + ... + 19 + 30 + 31 + ... + 39.
15. If we let BE vary, the configuration is uniquely determined by the angle APC. Guess that it is 90o. Take DX = 6. Then by similar triangles, Bx = 9. Area ACD = 54. Area ABX = 81, but area BDX = 27, so area ABC = 108. It remains to show that BE = 20. BF = 15 (3, 4, 5 triangle). Line BFE meets three sides of triangle ACD, so (-1/2)(1/1)(CE/EA) = -1. CE = 3 sqrt(13), so EA = sqrt(13), CE = 2 sqrt(13). Applying cosine formula to FEA and FEC, 81 = 52 + y2 - 2y sqrt(13) cos CEF, 36 = 13 + y2 + y sqrt(13) cos CEF. Hence 75 = 3y2, y = 5 and BE = 20 as claimed.
© John Scholes
jscholes@kalva.demon.co.uk
14 July 2003
Last updated/corrected 21 Dec 03