| 1 252, 1022 | 9 192 |
| 2 169 | 10 840 |
| 3 27 | 11 163 |
| 4 20 | 12 441 |
| 5 9/625 | 13 987 |
| 6 142 | 14 84 |
| 7 110 | 15 704 |
| 8 364 |
1. 10C5 = 252, 210 - 2.
2. We find f 4(11) = 169, f 6(11) = 169, hence result.
3. Put log2x = 3k, then we have k3 = 3k, so k = 31/2.
4. lhs < n, so we need n ≥ 20. Taking xi = alternately ±19/20 works for n = 20.
5. 104 factors, 122 divisible by 1088, because factors are 2m5n and we need m, n = 88, 89, ... , or 99.
6. Let a be the number next to the 0 as shown. Then A = 4a, so B = 93 + 2a, so C = 108 - a, so 74 = 324 - 5a, so a = 50. Hence D = 150, E = 148, so * = 142
7. Let altitude from A be length h. Using tan(x + y) = (tan x + tan y)/(1 - tan x tan y) we have 22/7 = (3/h + 17/h)/(1 - 51/h2). Hence 11h2 - 70h - 561 = 0. Hence h = 11.
8. f(14, 52) = 26/19 f(14, 38) = (26/19) (19/12) f(14, 24) = (26/12) (12/5) f(14,10) = (26/5) (7/2) f(10, 4) = (91/5) (5/3) f(4, 6) = (91/3) 3 f(4, 2) = 91·2 f(2, 2) = 91·2·2.
9. Let number be N. Only single digit cube ending in 8 is 23. So must be (10a + 2)3 = 120a + 8 mod 100. Only single digit multiples of 12 ending in 8 are 4·8 = 48 and 9·12 = 108, so a = 10b + 4 or 10b + 9, and N = 100b + 42 or 100b + 92. Hence N3 = 3·422100b + 423 or 3·422100b + 923 mod 1000 = 200b + 88 or 200b + 688 mod 1000. So smallest is 192.
10. There are (12·4 + 8·6 + 6·8)/3 = 48 vertices. From each vertex there are 3 edges and 1 + 3 + 5 = 9 diagonals. Hence 47 - 3 - 9 = 35 internal segments. Hence 48·35/2 in total.
11. Sum wm = 3 + 504i. Put zm = 3i + km(a + bi) with real km. Then sum zm = 15i + k(a + bi), where k = sum km. Hence b/a = 489/3 = 163.
12. Area PAB/area ABC = 3/(c+3). Hence 1 = 3/(a+3) + 3/(b+3) + 3/(c+3). Multiplying out we get abc = 9(a + b + c) + 54.
13. Starting at x0, we find successively that the coefficients of the other factor are -1, 1, -2, 3, -5, 8, -13, 21, -34, 55, -89, 144, -233, 377, -610, 987, which are Fibonacci numbers with alternating signs.
14. We first find the reflection (A, B) of the point (a, b) in the line y = 2x. The midpoint is ( (A+a)/2, (B+b)/2 ), so (B + b)/2 = A + a, or 2A - B = b - 2a. The slope of the line joining is -1/2, so (B - b) = -(A - a)/2, or A + 2B = a + 2b. Solving A = (-3a + 4b)/5, B = (4a + 3b)/5. So if AB = 1, then (-3a + 4b)(4a + 3b) = 25, or 12a2 - 7ab - 12b2 + 25 = 0.
15. 8 has been typed, so 1, 2, ... , 7 have already been put into the tray. Any left must be typed in decreasing order. 9 can come anywhere in the order or not at all (if already typed). So for a subset of {1, 2, ... , 7} with k elements, there are k+2 possibilities for 9. Hence 1·2 + 7·3 + 21·4 + 35·5 + 35·6 + 21·7 + 7·8 + 1·9 = 2 + 21 + 84 + 175 + 210 + 147 + 56 + 9 = 704.
© John Scholes
jscholes@kalva.demon.co.uk
13 July 2003
Last updated/corrected 13 Jul 03