By Pythagoras, AB = 25, AC = 15. Extend AC to X with AX = 25, then X must be (7,-15), so the midpoint of BX is (-4,-17), and the equation of the line joining it to A is (x+8)/(y-5) = -4/22, or 11x + 2y + 78 = 0. hence a = 11, c = 78.
| 1 528 | 9 9/64 |
| 2 828 | 10 144 |
| 3 117 | 11 23 |
| 4 13 | 12 144(2+√2+√3+√6) |
| 5 432 | 13 184 |
| 6 840 | 14 594 |
| 7 11,78 | 15 20 |
| 8 560 |
1. There are 22 squares in {1, 2, ... , 528}, 8 cubes and 2 sixth powers. Hence 528 - 22 - 8 + 2 = 500 which are neither squares nor cubes.
2. (3 + √43)2 = 9 + 6√43 + 43, so expr = (3 + √43)3 - (√43 - 3)3 (taking both square roots as positive) = 2(33 + 3·3·43) = 828.
3. Angles are 180o(1 - 2/r), 180o(1 - 2/s), so 58(r-2)s = 59r(s-2), or r = 116s/(118-s). But r is positive, so s ≤ 117. That works (r = 116·117).
4. Put y = x2-10x-29, so 1/y + 1/(y-16) = 2/(y-40), giving y = 10. Hence (x+3)(x-13) = 0, so x = 13.
5. 75 = 3·5·5 = (2+1)(4+1)(4+1), so n must be p2q4r4 for some primes p, q, r. To make n as small as possible we take 243452, which is divisible by 75. Hence n/75 = 2433 = 432.
6. Suppose initially N fish. Then after 6 mos, N x 75%/60% = 5N/4 fish, and 60 x 75% = 45 tagged. Estimate 45/(5N/4) = 3/70, or N = 840.
7. AB has gradient (5+19)/(15-8) = 24/7, AC has gradient (19-7)/(15+1) = 3/4
By Pythagoras, AB = 25, AC = 15. Extend AC to X with AX = 25, then X must be (7,-15), so the midpoint of BX is (-4,-17), and the equation of the line joining it to A is (x+8)/(y-5) = -4/22, or 11x + 2y + 78 = 0. hence a = 11, c = 78.
8. Label the columns 1, 2, 3. An order is a string of three 1s, two 2s and three 3s. There are 8!/(3!2!3!) = 560 such strings.
9. Let an be the number of strings of H and T of length n with no two adjacent Hs. Then a1 = 2, a2 = 3. We find an+2 = an+1 + an, because the string must begin T or HT. So we have the Fibonacci sequence: 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 and prob = 144/1024 = 9/64.
10. Put k = e2πi/144, then A = {k8, k16, k24, ... k144}, B = {k3, k6, k9, ... , k144}. Clearly any zw has the form kn, so there are at most 144 distinct products. 2·8 + 43·3 = 145 = 1 mod 144, so k is a product zw. Hence so also is kn = znwn.
11. The smallest product of n-3 consecutive positive integers is (n-3)! That is always too small. Similarly (n-2)!, (n-1)!/2 and n!/6 are too small. The next smallest is (n+1)!/24 = n! (n+1)/24. For n+1 > 24 that is too big and hence there are no such products for n > 23. For n = 23 it works.
12. The 12 sides have length 24 sin 15o. Similarly, there are 12 diagonals each of length 24 sin 30o, 24 sin 45o, 24 sin 60o, 24 sin 75o and 6 of length 24. So the total length is 144 + 288(sin 15o + sin 30o + sin 45o + sin 60o + sin 75o). Now cos 30o = cos(45o - 15o) = cos 45o cos 15o + sin 45o sin 15o = (sin 15o + sin 75o)/√2, so sin 15o + sin 75o = (√6)/2. Hence sin 15o + sin 30o + ... + sin 75o = (√6)/2 + (√3)/2 + 1/2 + 1/√2 = (1 + √2 + √3 + √6)/2. Hence total length 144(2+√2+√3+√6).
13. If 9n starts with 9, then 9n-1 must start with 1 and have the same number of digits. If 9n starts with any other digit, then it has one more digit than 9n-1. Now 94000 has 3816 more digits than 91. So in 3816 cases from 2, 3, ... , 4000, the first digit of 9n is not 9 and in the other 183 cases it is 9. Add in the case 91 (but not 94000) and we get 184.
14. I cannot see any elegant solution to this. It seems to be a horrible slog. Take the base ADC to have sides m, n, n where m = √432, n = √507. Then the three sides from the base to the apex O are all equal to ½√(m2 + n2). So if the foot of the altitude from O is X, then OXA, OXC, OXD are congruent, so X is the circumcenter of ADC.
By similar triangles CE/AC = CD/CY, so CE = n2/√(n2 - m2/4). Hence OX2 = (m2 + n2)/4 - n4/(4n2 - m2) = ½m2(3n2-m2)/(4n2-m2). Area ADC = ½m√(n2 - m2/4), so vol = (1/3) OX area ADC = m2/24 √(3n2-m2) = 594.
15. This needs a trick: (axn+1 + byn+1)(x+y) - (axn + byn)xy = axn+2 + byn+2. Hence 7(x+y) - 3xy = 16, 16(x+y) - 7xy = 42. Solving x+y = -14, xy = -38. Applying again: ax5 + by5 = 42(x+y) - 16xy = 20.
© John Scholes
jscholes@kalva.demon.co.uk
6 Oct 2003
Last updated/corrected 6 Oct 03