| 1 336 | 9 6 |
| 2 120 | 10 48 |
| 3 192 | 11 (527/40)√11 |
| 4 1/27 | 12 134 |
| 5 216π | 13 171/512 |
| 6 112 | 14 48√3 |
| 7 400 | 15 8 + 4√3 |
| 8 348 |
1. Let the numbers be a, b, a+b. So ab(a+b) = 12(a+b). Hence ab = 12. So {a, b} = {1, 12}, {2, 6} or {3, 4}. Hence N = 156, 96 or 84, sum 336.
2. The best we can hope for is to use each digit just once, and for the larger digits to occur earlier. Any multiple of 1000 is divisible by 8, so 9876543000 is a multiple of 8. Now 210 and 201 are not multiples of 8. The next best is 120. So the number is 9876543120, with residue 120.
3. We have 3 choices for the first letter, and 2 for each subsequent letter, so 3·26.
4. The centroid of each face is 1/3 of the way up the median, so the top face of T' is 1/3 of the way up the height of T. T' is obviously regular, so it is 1/3 the scale of T, and its vol is 1/27 of T.
5. Take another plane cut parallel to the first and with its intersection to the second parallel to the intersection of the first two and also meeting the log at a single point. Then the two wedges are congruent, and together they form a cylinder diameter 12, length 12 and hence vol 12·π62.
6. The overlap is a hexagon, and there are six congruent triangles outside the hexagon, each similar to the original triangle and 1/3 the (linear) size (because the centroid divides the median in the ratio 2:1). So the area of the union is 4/3 times the area of the triangle. We have the semiperimeter s = 21, so by Heron's formula the triangle has area √(21·8·7·6) = 84.
7. AD = 2 AO cos x and = 2 AO' cos y = 2 AO' sin x = 4 AO sin x, so cos x = 2 sin x, so tan x = 1/2, cos x = 2/√5. Area rhombus = 2 AD2 sin x cos x = 25 cos4x = 400.
8. (A-D)(a-d) = 1440, Aa = 1716, (A+D)(a+d) = 1848, so 2Dd = 1440 + 1848 - 2·1716 = -144. Hence (Ad + Da) = 204, so 8th term (A+6D)(a+6d) = Aa + 36Dd + 6(Ad+Da) = 1716 - 36·72 + 6·204 = 348.
9. x4-x3-x2-1 = (x+1)(x3-2x2+x-1), so a = -1, b+c+d = 2, bc+cd+db = 1. Hence b2+c2+d2 = 22-2 = 2. We have p(x) = (x3-2x2+x-1)(x3+x2+x+1)+x2-x+1. Hence p(a) = 3, p(b) = b2-b+1, p(c) = c2-c+1, p(d) = d2-d+1, so p(a)+p(b)+p(c)+p(d) = 6 + (b2+c2+d2)-(b+c+d) = 6.
10. Squaring, we see that n(n+60) must be a square. If either of n, n+60 is a square, then the other is and hence m is a square. Contradiction. So neither n nor n+60 is a square. Thus we must have n = a2d, n+60 = b2d for some non-square d. So 60 = d(b2-a2). b+a and b-a have the same parity, so b2-a2 must be odd or a multiple of 4. Hence d = 3, 5 or 15. This leads to n = 20 or 48 as the only solutions.
11. Extend AC and MD to meet at X. Let Y be the foot of the perpendicular from C to DM. XMA and BCA are similar, so XA = 625/14. Hence CY/AM = XC/XA = 527/625, and CY = 527/50. DM = (5/2)√11, so area CDM = (527/40)√11.
12. n = 135 works with 5 votes per candidate. Then each has 5/135 = 3.7%. No candidate can have only 1 vote. If a candidate has just 2 votes, then 2/n <= 1/100, so n >= 200. If a candidate has 3 votes, then n >= 150. So in a minimal solution each candidate must have at least 4 votes. If all have at least 5, then n >= 135. If a candidate has 4, then 4/n <= 3/100, so n >= 134. This can be achieved: 1 candidate has 4 votes, the other 26 have 5 each. Then 5/134 = 3.7%, 4/134 = 2.99%.
13. Label vertices 0, 1, 2. Suppose A moves +1 mod 3 and B moves +2 mod 3. Then A + 2B = 0 mod 3 and A + B = 10. Hence A = B = 2 mod 3. So (A, B) = (2, 8), (5, 5) or (8, 2). There are 10C2, 5C5, 10C8 possibilities of each type. Hence 45 + 252 + 45 = 342 in all. So prob = 171/512.
14. wlog B is lower than F, so the y-coords of B, C, D, E, F must be 2, 6, 10, 8, 4 respectively. Angle A = 120o, so y = 60o - x. B is half the height of F, so sin(60o-x) = 2 sin x, giving tan x = (√3)/5 and side AB = 4√(7/3) = k. E is twice the height of F and FE = FA, so E must be vertically above A. Similarly, D is vertically above B. Area AFE = area BCD = 4k cos y. Area ABDE = 8k cos x. So area hexagon = 8k(cos x + cos y) = 4k(3 cos x + √3 sin x) = 48√3.
15. The polynomial factorises as x(x23 + x22 + ... + 1)2, with distinct roots x = 0, and e2πki/24 for k = 1, 2, .. , 23. We can ignore x = 0, whose square is real. The square of e2πki/24 is e2πki/12 with imaginary part sin(πk/6). So the sum is 4 + 8 sin(π/6) + 8 sin(&pi/3) (and 3 zero terms) = 8 + 4√3.
© John Scholes
jscholes@kalva.demon.co.uk
4 Aug 2003
Last updated/corrected 26 Nov 03