| 1 217 | 9 3/32 |
| 2 201 | 10 375/442 |
| 3 241 | 11 125/387 |
| 4 858 | 12 5/9 |
| 5 349/500 | 13 159/323 |
| 6 882 | 14 (60 - √750)/3 |
| 7 -588 | 15 29509 |
| 8 199 |
1. n = 9876 = 34 mod 37, or 8765 = 33 mod 37, or 7654 = 32 mod 37, or 6543 = 31 mod 37, or 5432 = 30 mod 37, or 4321 = 29 mod 37, or 3210 = 28 mod 37. So sum = 34 + 33 + 32 + 31 + 30 + 29 + 28 = 217.
2. (a+1) + ... + (a+m) = m(2a+m+1)/2 = 2m, and (b+1) + ... + (b+2m) = m(2b+2m+1) = m. Also b+m-a = 99. Solving a = -99, b = -201, m = 201.
3. Label vertices i = 1 to 26. Let i belong to ti triangles and qi quadrilaterals. Then i belongs to (25-ti-2qi) space diagonals. So total = ½∑(25-ti-2qi) = 325 - ½∑ ti - ∑ qi = 325 - 72/2 - 48 = 241.
4. If ER = 2, and O is the midpoint of ER, then O is the center of the circle through E,Q,R, so OQ = 1. Hence O lies on the quarter circle center Q radius 1. Thus the midpoints enclose the region shown in the second diagram.
So A = 4 - π, and 100A is about 858.
5. A's ratio was 8/15 on day 1, 7/10 on day 2, and 8/15 < 7/10. We want B's ratio to be weighted towards day 2, so we take 1/2 on day 1 and 348/498 on day 2 (we must have a total of 500 attempted), giving 349/500 overall.
6. Given digits a < b < c < d, the possible snakelike numbers are acbd, adbc, bcad, bdac, cdab, unless a = 0, in which case only the last three are possible. There are 9C4 = 126 ways of choosing 4 non-zero digits, and 9C3 = 84 ways of choosing 3 non-zero digits, so 5·126 + 3·84 = 882 snakelike 4-digit nos.
7. -1+2-3+4-...+14-15 = -8. So twice coeff is -1(-8+1) + 2(-8-2) - 3(-8+3) + ... -15(-8+15) = (-1+2-3+...-15)(-8) - (12+22+...+152) = 64 - 15·16·31/6 = -1176, and coeff is -588.
8. Evidently Pi must be the vertices of a regular n-gon (but in a different order). So we want number of m st 2 ≤ m ≤ 499 with m coprime to 1000, in other words odd and a not a multiple of 5. There 249 odd nos. 3, 5, 7, ... , 499, of which 50 are multiples of 5: 5, 15, ... , 495. Hence 199 non-similar stars.
9. The line dividing the rectangle must pass through a vertex, so take it to be DZ.
If D is not the smallest angle of DZG, then GZ ≥ 4/3 DG. Contradiction. So D is the smallest angle, so either DG = 6 and GZ = 4½, or DG = 7 and GZ = 5¼. We want area AXY to be as small as possible, so BX/BC to be as big as possible and hence EF/DE to be as big as possible. So we take EF = 7. Hence AX = 1/2, XY = 3/8, area AXY = 3/32.
10. If the circle lies entirely inside the rectangle, then its center must lie inside a 13x34 rectangle, area 442. If it does not intersect the diagonal, then its center must lie inside one of the two triangles shown. Note that a 5, 12, 13 triangle has inradius 2 (because if the inradius is r, then calculating the area two ways we get 5·12/2 = r(5+12+13)/2). So the triangles are 12½,30, 32½. Thus their area is 30·12½ = 375. So prob = 375/442
11. Let small cone have height h. Then vol cone/vol frustrum = h3/(64-h3). If we unroll large cone, it forms part of circle with radius 5, arc length 6π, so area (6π/10π)25π = 15π. Base has area 9π. So painted area of small cone is h215π/16, painted area of frustrum = 9π + (16-h2)15π/16. Thus k = h3/(64-h3) = 5h2/(128-5h2). So h = 5/2, k = 125/387.
12. We must have x ∈ (1/2, 1] ∪ (1/8, 1/4] ∪ (1/32, 1/16] ∪ ... and y ∈ (1/5, 1] ∪ (1/125, 1/25] ∪ (1/3125, 1/625] ∪ ... So we have a set of rectangles with total area (1/2 + 1/8 + 1/32 + ... )(4/5 + 4/125 + 4/3125 + ... ) = 1/2(1 + 1/4 + 1/42 + ... )4/5(1 + 1/25 + 1/252 + ... ) = (2/5)(1/(1 - 1/4))(1/(1 - 1/25)) = (2/5)(4/3)(25/24) = 5/9.
13. We have x36 - 2x18 + 1 = x19 - 2x18 + x17, so (x19 - 1)(x17 - 1) = 0. We exclude the two roots x = 1 (which we introduced by multiplying by (x-1)2). So the ai are 1/19, 2/19, ... , 18/19 and 1/17, 2/17, ... , 16/17. Thus sum = 1/19 + 2/19 + 3/19 + 1/17 + 2/17 = 6/19 + 3/17 = 159/323.
14. The first diagram shows a horizontal plane through the point where the rope is attached to the unicorn. Evidently the tangent is length 4√5.
Unroll the surface of the tower. We have AC = 20, CD = 4√5, DE = 4, so AB = (8√6 - 4√5)20/(8√6) = (60 - √750)/3.
15. Most integers have two possible predecessors. Let M be the operation of moving back from n to n-1 and D the operation of moving back from n to 10n. There are two related difficulties. We cannot take M10 on a multiple of 10 (because M does not apply to multiples of 10 plus 1); we cannot even take M9 on 10 (because having reached 1 we must stop). The predecessor of 1 must be 10. Then we need a further 18 moves back. So, leaving aside the difficulties above, there are 218 possibilities. We must exclude the 29 M9X...X, where X denotes M or D. Then we must exclude the 27 DM10X...X. Similarly, we must exclude the 27 XDM10X...X, the 27 XXDM10X...X, ... , and the 27 X...XDM10. Thus we are left with 218 - 29 - 8·27 = 218 - 3·29 = 29509.
© John Scholes
jscholes@kalva.demon.co.uk
25 March 2004
Last updated/corrected 25 Mar 04