| 1 200 | 9 342 |
| 2 340 | 10 159 |
| 3 44 | 11 276 |
| 4 1/6 | 12 55/3 |
| 5 23 | 13 27/38 |
| 6 17/32 | 14 768 |
| 7 300 | 15 7/9 |
| 8 799 |
1. We find successively, a = 114, b = x - 95, c = 191, d = x - 190. Then 115 = b + d = 2x - 285, so x = 200.
2. [log2n] = 2 for n = 4, 5, 6, 7; 4 for n = 16, 17, ... , 31; 6 for n = 64, 65, ... , 127; 8 for n = 256, 257, ... , 511. So 4 + 16 + 64 + 256 = 340.
3. There are potentially (n+1)2 terms, namely xayb for 0 ≤ a, b ≤ n. So to get at least 1996 we need n = 44. The only problem is whether some of these terms have zero coefficient. But if (xy)axbyc = xyAxByC, then a+c = A+C. Signs are (-1)b(-1)n-a-b-c = (-1)n-a-c, and (-1)n-A-C, so they are equal. So no terms vanish.
4. By similar triangles the total shadow (including the part under the cube) is a square with side (1+x)/x. So (1+x)2/x2 = 49, or (8x+1)(6x-1) = 0.
5. a+b+c = -3, so the required equation has roots -3-a, -3-b, -3-c. If y = -3-x, then (-3-y)3 + 3(-3-y)2 + 4(-3-y) - 11 = 0, or y3 + ... + 27-27+12+11, so t = 23.
6. Prob team A wins all 4 games is 1/16. Two teams cannot both win all their games, so prob some team wins all 4 games is 5/16. Similarly prob some team loses all 4. But these overlap. Prob team A wins all 4 games, and team B loses all 4 games is 1/128 (7 games determined). So prob some team wins all 4 and some team loses all 4 is 20/128 = 5/32. So prob some team wins all 4 or some team loses all 4 is 10/16 - 5/32 = 15/32. Hence requd prob is 17/32.
7. There are 49·48/2 = 1176 ways of choosing 2 cells. In 24 cases the two cells are diametrically opposite. In the other 1152 they are not. The 1152 form groups of 4 related by rotation, so 288 distinct. The 24 form 12 groups of 2. So total 300.
8. We have mn = 219320m + 219320n, or (m - 219320)(n - 219320) = 238340. Now 238340 has 39·41 factors. Leaving aside the square root 219320 which does not give a solution, the factors form 799 pairs (one < square root and one >). Each pair gives a solution m, n.
9. Write the labels in binary. 1st pass, open those ending in 1, leaving those ending in 0 still closed. 2nd pass, open those ending in 00, leaving -10 closed. 3rd pass open 010, leaving -110 closed. 4th pass open -1110, leaving -0110 closed. 5th pass, open -00110, leaving -10110. 6th pass open -110110, leaving -010110. Continuing, we leave the single locker 0101010110 = 342.
10. (cos 96o + sin 96o)/(cos 96o - sin 96o) = (1 + tan 96o)/(1 - tan 96o). Now we need a trick, which is that tan 45o = 1, so we can write the expression as (tan 45o + tan 96o)/(1 - tan 45otan 96o) = tan 141o. Now tan 19no = tan 141o iff 19n = 141 mod 180. We have 180 = 9·19+9, 19 = 2·9 + 1. So 1 = 19 - 2·9 = 19 - 2(180 - 9·19) = 19·19 - 2·180. So 141 = (19·141)19 - 282·180, or 141 = 19(19·141) mod 180. But 19·141 = 14·180 + 159, so 141 = 19·159 mod 180.
11. We need the factorisation: z6 + z4 + z3 + z2 + 1 = (z4 + z3 + z2 + z + 1)(z2 - z + 1). So the roots are ei72o, ei144o, ei216o, ei288o, ei60o, ei300o. The roots with positive imaginary part are ei72o, ei144o, ei60o, with product ei276o.
12. By symmetry the average of |ai - aj| is independent of i, j (provided they are unequal). Suppose ai = k. Then the average of |k - aj| is (k-1 + k-2 + ... + 1 + 1 + 2 + ... + 10-k)/9 = (k2 -11k + 55)/9. So average of |ai - aj| is (1/10) ∑ (k2 -11k + 55)/9. Hence requd average is ∑ (k2 -11k + 55)/18 = (385 - 605 + 550)/18 = 55/3.
13. We have area ADB/area ABC = area ADB/(2 area ABM) = AD/(2 AM).
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© John Scholes
jscholes@kalva.demon.co.uk
4 Oct 2003
Last updated/corrected 4 Oct 03