| 1 630 | 9 16/45 |
| 2 651 | 10 23/177 |
| 3 500 | 11 149 |
| 4 216+72√3 | 12 2/3 |
| 5 (16√3)/13 | 13 -9+√165 |
| 6 7/72 | 14 351 |
| 7 860/63 | 15 1/84 |
| 8 315 |
1. Obv 11, 22, ... , 99 sum 11·45 = 495 are solutions. ab must be divisible by a, so only other possibilities are a(2a), a(3a) etc. Hence none for second digit ≥ 5. Easily find 48 works, 36 works, 39 does not. 24 works, 26, 28 do not. 12, 15 work. So other solns sum 12+15+24+36+48 = 135.
2. Suppose S has n els, total N. Then (N+1)/(n+1) = N/n - 13, or N = 13n2+14n. Also (N+2001)/(n+1) = N/n + 27, or 1974n-27n2 = N. Hence n = 49, N/n = 651.
3. Expanding, coeff x2001 = 0, coeff x2000 = 2001/2, coeff x1999 = -250·2001. Hence sum = 500.
4. The trick is that ATC is isosceles, because ∠C = ∠T = 75o. Now take CS as an altitude. Then CS = AC sin 60o = 12√3 and AS = 12. Also ∠B = ∠SCB = 45o, so SB = 12√3. Hence AB = 12+12√3. So area = AB·CS/2 = (6+6√3)12√3 = 216+72√3.
5. The line through (0,1) and slope -√3 is (y-1) = -x√3. This intersects the ellipse at x2 + 4(1-x√3)2 = 4, or 13x2 - 8x√3 = 0, so x = (8√3)/13, y = -11/13. Side length is 2x (because other vertex is (-x,y) ) = (16√3)/13
6. Given the pattern of values (one 6, two 5s, one 3, for example), there is unique admissible sequence of throws giving those values (eg 3, 5, 5, 6). The no. of patterns is the same as the number of ways of placing 4 identical balls in 6 urns. That is the same as the number of ways of placing 4 balls and 5 dividers in a row, or 9C4 = 126. Hence prob = 126/64 = 7/72.
7. Let area be A, inradius be r, height from base 20 be h. Then A = ½h20 = ½r(20+21+22), so r/h = 20/63. By similar triangles XY = 20(h-r)/h = 20(1-r/h) = 20·43/63.
8. a0+10a1+102a2 + ... = 2(a0+7a1+72a2+...) or a0 + 4a1 = 2a2 + 314a3 + 7599a4 + ... . But ai ≤ 6, so lhs ≤ 30, so a3, a4, ... = 0 and a0 + 4a1 = 2a2. Largest soln is evidently a2 = 6, a1 = 3, a0 = 0, giving 6307 = 31510 (note that ai must be ≤ 6).
9. area ADC = α area ABC. area ADF = (1-γ)area ADC = α(1-γ) area ABC. Similarly for the other two triangles, so area DEF/area ABC = 1 - (α+β+γ) + (αβ+βγ+γα) = 1/3 + ½( (α+β+γ)2 - (α2+β2+γ2) ) = 1/3 + 1/45 = 16/45.
10. The midpoint of (x,y,z) and (x',y',z') lies in S iff x,x' have the same parity, y,y' have the same parity, and z,z' have the same parity. There are 22+12=5 ways of choosing x,x' to have the same parity, incl 3 with x=x'. Similarly, 8 incl 4 for y,y' and 13 incl 5 for z,z'. Hence 5·8·13 - 3·4·5 = 460 of choosing distinct points with same parity. There are 60·59 ways of choosing distinct points, so prob 460/3540 = 23/177.
11. Suppose the row 1 point is in col a, the row 2 point in col b and so on. Then a = 5b-3, b+N = 5a-4, c+2N = 5d-1, d+3N = 5e, e+4N = 5c-2. First two equs give a = 5b-3, N = 24b-19. Last three give 124e = 89N+7. Hence 124e = 2136b-1684. Now 2136b-1684 = 28b+52 mod 124, and by trial smallest possible solution (for b and hence for N) is b = 7, giving N = 149, e = 107. Continuing, we get a = 32, d = 88, c = 141. So it is indeed a solution.
12. We use the formula r x total face area = 3 vol. Triangle (0,0,0), (6,0,0), (0,4,0) has area 12. Hence vol = 2·12/3 = 8. Triangle (6,0,0), (0,4,0), (0,0,2) has sides √20, √40, √52. Slogging out Heron gives area 14. Hence total face area = 14 + 12 + 6 + 4 = 36. Hence r = 24/36 = 2/3
13. Let radius be r, k = d/2, then 2r sin k = 22, 2r sin 2k = x + 20, 2r sin 3k = x. So (sin 2k)/sin k = (x+20)/22, (sin 3k)/sin k = x/22. But (sin 2k)/sin k = 2 cos k, (sin 3k)/sin k = 4 cos2k - 1. Hence x/22 = ( (x+20)/22)2 - 1, or x2 + 18x - 84 = 0. Must take positive root, so x = √165 - 9.
14. Let an, bn, cn be no. of sequences length n beginning with 1, 01, 001 respectively. Then an+1 = bn + cn, bn+1 = an, cn+1 = bn. So an+1 = an-1 + bn-1, bn+1 = bn-1 + cn-1, cn+1 = an-1. Put dn = an + bn + cn. So dn+1 = dn-1 + an-1 + bn-1 = dn-1 + dn-2. Note that d1 = 2 (1, 0), d2 = 3 (10, 01, 00), d3 = 4 (101, 100, 010, 001). So we now calculate successively 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351.
15.
© John Scholes
jscholes@kalva.demon.co.uk
10 October 2003
Last updated/corrected 23 Mar 04