| 1 9 | 9 28501 |
| 2 294 | 10 360π/(180-π),180π/(180+π) |
| 3 111 | 11 (√5 - 1)/8 |
| 4 (361803√3)/2 | 12 243623/510 |
| 5 42 | 13 225/676 |
| 6 521 | 14 95/3 |
| 7 112 | 15 (12√221)/49 |
| 8 49 |
1. n = abc, with 1 ≤ a,c ≤ 9. So |n-n'| = 99|a-c|. There are 9 possible values for |a-c|, namely 0, 1, ... , 8.
2. We find PQ = QR = PR = √98. If a cube has side a and three vertices form an equilateral triangle, then the triangle must have side a√2, so a = 7. So area = 6a2 = 294.
3. log6a + log6b + log6c = 6 is equivalent to abc = 66. a, b, c GP is equivalent to b2 = ac, so b3 = 66, and b = 36. b-a is square, so a = 11, 20, 27, 32, 35. But a divides 64, so a = 27. Hence c = 48.
4. A large hexagon with n unit hexagons on a side can be divided into 1 + 6(1 + 2 + ... n-1) = 3n2-3n+1 unit hexagons. So for n=202 the hole contains 3·2012 - 3·201 + 1 = 120601 unit hexagons, each area (3√3)/2. Total area (361803√3)/2
5. We must have either 6a > 2a3b, or 6b > 2a3b. If b = 0, we must have 6a > 2a, so a = 1, 2, 3, 4. If a = 0, we must have 6b > 3b, so b = 1, 2. Suppose b > 0 and a > 0. If 6a > 2a3b, then 6a > 2a3, so 2a > 2a, which is not possible. Similarly, if 6b > 2a3b, then 3b > 3b, which is not possible. So only solutions 2, 4, 8, 16, 3, 9.
6. Expr = 250 ∑(1/n-2 - 1/n+2) = 250(1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 - 1/10002) = 520 + 5/6 - 250(1/9999 + 1/10000 + 1/10001 + 1/10002). The last term is about -1/10 and certainly > -2/6, so nearest integer 521.
7. We want n(n+1)(2n+1) a multiple of 1200 = 243·52. It is always divisible by 3, so we need it divisible by 16 and 25. n, n+1, 2n+1 are relatively prime. So (A) 16|n, 25|n+1, (B) 16|n, 25|2n+1, (C) 16|n+1, 25|2n+1, (D) 25|n, 16|n+1. If (A), put n = 16k, so 16k+1 is multiple of 25. Hence also 16k-24 = 8(2k-3), so smallest k = 14, n = 224. If (B), put n = 16k, so 32k+1 is multiple of 25, so 7k+1 is multiple of 25, smallest k = 7, n = 112. If (C), put n=16k-1, so 32k-1 is multiple of 25, hence also 7k-1. Smallest k = 18, n = 287. If (D), put n = 25k, so 25k+1 is multiple of 16, hence also 7k-1. Smallest k = 7, n = 175.
8. Suppose we gradually reduce x from 2002. As we go from k+1 to k, 2002/x is increased by 2002/(k(k+1)). Since 44·45 = 1980 < 2002, this increase is < 1 whilst x remains above 44. If we reduce h by < 1, then we cannot reduce [h] by more than 1. So we cannot miss values whilst x ≥ 44. [2002/44] = 45, so we get 1, 2, ... , 45. [2002/43] = 46, [2002/42] = 47, [2002/41] = 48, [2002/40] = 50. So first missed is 49.
9. To get A, B disjoint we must place each element in just A, just B or neither, 3 choices, or 3n in all (for subsets of {1, 2, ... , n}). That counts all except the empty set twice, so there are (3n-1)/2 possibilities = 29524. But that includes the 2n-1 = 1023 possibilities where one subset is empty (and the other not). Thus there are 29524 - 1023 = 28501.
10. The smallest value x satisfies 180x/π = 180 - x or x = 180π/(180+π) degrees The next smallest satisfies 180x/π = 360 + x or x = 360π/(180-π).
11. a + ak + ak2 + ... = a/(1-k). So a = 1-k and |k| < 1. If we have another sequence (1-h), (1-h)h, (1-h)h2 ... with same second term, then (1-k)k2 = (1-h)h2 implies h-k=0 or h+k=1. h=k gives same sequence, so must have h = 1-k. Must have k ≠ ½ or that also gives same. Third term = 1/8 gives 8k3 - 8k + 1 = 0, or (2k-1)(4k2-2k-1) = 0. Hence 4k2-2k-1=0. So k = (1+√5)/4 or (1-√5)/4. Corresponding h=1-k are (3-√5)/4 and (3+√5)/4. Latter is > 1. So k = (1+√5)/4 and second term (√5-1)/8.
12. Total Hs must be 0 after tosses 1, 2; ≤1 after 3, 4; ≤2 after 5, 6, 7; ≤3 after 8; 3 after 9; 4 after 10. Thus the sequence of throws must be one of these 23:
1 2 3 4 5 6 7 8 9 10 T T H T H T T H T H T T H T H T T T H H T T H T T H T H T H T T H T T H T T H H T T H T T T H H T H T T H T T T H T H H T T H T T T T H H H T T T H 7 poss as above T T T T H H T H T H T T T T H H T T H H T T T T H T H H T H T T T T H T H T H H T T T T H T T H H H T T T T T H H H T H T T T T T H H T H H T T T T T H T H H H T T T T T T H H H HSo prob. 23(0.4)4(0.6)6 = 23·2436/510.
13. Take line through E parallel to AD. Suppose it meets CD at F. Then CF/CD = CE/CA = 1/4, so CF = 1/2. Hence BP/BE = BD/BF = 5/(6.5) = 10/13. So PQ/EA = 10/13 also (triangles BPQ, BEA). Hence PQ = 30/13. PQR and CAB are similar, so area PQR/area ABC = (PQ/CA)2 = (30/52)2 = (15/26)2
14. Put OP = x, OPT has sides x, 19, √(x2-361). It is similar to MPA which therefore has perimeter (19 + x + √(x2-361)) (19+x)/√(x2-361) = 152. Put x = 19y. Dividing through by 19: (1 + y + √(y2-1) ) (1+y)/√(y2-1) = 8. Rearranging: (7-y)√(y2-1) = (1+y)2. Squaring etc, 9y2 - 30y + 25 = 0, or (3y-5) = 0, so y = 5/3, x = 95/3.
15. Suppose centers lie on y = nx. Then using tan 2k = 2 tan k/(1-tan2k), we have m = 2n/(1-n2). If the center is (a,na), then radius must be na. So (a-9)2 + (na-6)2 = (na)2, or a2 - (18+12n)a + 117 = 0. So the product of the two possible values of a is 117. Hence the product of the two radii is 117n2 = 68. Hence m = 2n/(1-n2) = (12√221)/49.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 2003
Last updated/corrected 10 Oct 03