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This puzzle will necessarily have several possible answers; here is one I came up with. To reach 9 words with only 10 different letters, each word in the list should only introduce one new letter, with the exception of the first word, although even that needs to be limited to just two different letters.
MAMMA MA
MAGMA G
LLAMA L
HALAL H
LAUGH U
GHOUL O
AMIGO I
HUMAN N
LOGIC C
10609 different ways. This is 103 squared.
Trying to go ahead and systematically list all of the ways is going to be a massive task, but luckily there is a quicker way.
The first go can either be 1, 2 or 3. If it was 1 then there would be 15 counters left; if it was 2 there would be 14 counters left; if it was 3, 13 left. All quite obvious so far. So if we only knew how many ways there were of combining to make 13, 14, and 15, we could just add them together to find the total number of ways for 16. We can do this safely knowing there is no overlap but that they cover all cases, as they all start with different numbers. All very nice, but the trouble is we don’t know how many ways there are of making 13, 14 or 15. Luckily our rationale will work in general, and for any number of board spaces, we only have to add up the previous three to find the answer. This holds until the number of counters is less than four, but luckily those numbers are more manageable.
1:1 way
2:2 ways (1+1 or 2)
3:4 ways (1+1+1, 1+2, 2+1, 3)
Building on this, knowing that each subsequent number of ways is simply the sum of the previous three, we can soon find the answer:
4:7
5:13
6:24
7:44
8:81
9:149
10:274
11:504
12:927
13:1705
14:3136
15:5768
16:10609
There is even a clever way to avoid working out the first few, that works in general. For each negative number there are 0 ways, and for zero there is 1 way, then every subsequent number of ways is the sum of the previous n. So for instance if n=6, ie. if you had a six-sided dice and wanted to know how many different ways of reach exactly 10 with multiple throws:
0:1
1:1
2:2
3:4
4:8
5:16
6:32
7:63
8:125
9:248
10:492
As an extra bonus question, does this sequence, the one based on a six-sided dice, ever yield a prime number of ways (apart for the 2 ways of getting 2)?
The squares have area 4, 9 and 31.25 respectively, and the triangles have areas of 2.25, 1 and 2.5. The total area is 50 square units.
Within each block, countries can be read diagonally downwards, jumping back to the top after the bottom row of the block. The message reads: YOU ARE OK.
They are 64, 61 and 59.
An even number of years must have past between the two scenarios, since at least one of the sisters must have had an odd age in the earlier scenario (there is only one even prime). So therefore the cube numbers must be of the same parity. They can’t both be odd, since 1 is not bigger than any prime numbers, and 125 exceeds a human lifetime, so 27 is the only viable odd cube. So the older sister must have been 8 in the early scenario and 64 now.
There are four prime number below 8: 2,3,5 and 7. Adding 56 years to these gives 58, 59, 61 and 63. Only 59 and 61 are prime, so these are the younger sisters’ ages now.
Staring with the central box, notice that 9 must be formed in different ways in both the vertical and horizontal directions. There are only three ways this is possible using the numbers 1 to 9 without repetition: 1,2,6; 1,3,5; 2,3,4. Since we know 2 and 3 are not in the same column or row, we know 1 must be in the same column and row as 2 or 3. Column 4 already has a 1, so the 1 can be placed in column 6, row 5. To complete the sums to 9, the 5 and 6 can be placed. By attempting to form the sums to 15 and 21 you will find there is only one way to place the remaining numbers in the central box. The next leap of insight is that for every horizontal sub-row, the 9 must be 1+2+6, the 15 3+4+8 and the 21 5+7+9. And for each vertical sub-column, 9 = 1+3+5, 15 = 2+4+9, 21 = 6+7+8. Any attempt to deviate from this will quickly lead to repetition of digits. It’s still not straightforward, but it is at least possible to logically deduce the remaining cells, broadly starting with the top middle box and moving anti-clockwise.
x is equal to 32 feet, such that the corridor is a little over 4 feet wide.
Lines F, H, N and O are all diameters of circles which pass through midpoints of sides of the original hendecagon.
The key insight is that an angle within a semicircle will be a right angle, so we start off with two construction lines at right angles, one collinear with one of the sides of the 11-sided polygon, another passing through the vertex opposite that side. By definition they will intersect at the midpoint of the side of the polygon.
On the left hand side draw extensions from all possible polygon edges that can meet the blue (horizontal) line, and on the right hand side draw extensions from all possible polygon edges that can meet the red (vertical) line (some will go upwards, others downwards). The red intersections are intersection points of extended polygon sides, as the red lines can be mirrored by the red axis. The blue intersections are already intersections of extended polygon sides as the blue axis itself is an extended polygon side.
The intersection points with the main axes can be paired up according to how far they are from the centre of the polygon. By definition any circle with one of these magenta pairing lines as its diameter will pass through the point where the axes meet (as well as a second polygon midpoint which will vary from circle to circle).
None of these pairing lines are horizontal as in the puzzle, but it’s easy to see how for each pairing line, the figure can be rotated such that the line is horizontal. If you do so you will find that the four lines correspond to the lines F, H, N and O in the puzzle.
The simplest way to calculate the areas of the rectangles is probably to complete the circles, with three additional copies of each quadrant, then draw a diameter between two opposite points as shown:
In the first case the area of the required rectangle is 2x^2, and in the second it is 2y^2.
In the first diagram, there is a right triangle with legs 2x and 4x, and hypotenuse 2.
(2x)^2 + (4x)^2 = 2^2
4x^2 + 16x^2 = 4
20x^2 = 4
2x^2 = 4/10 = 2/5
Similarly in the second diagram:
y^2 + (5y)^2 = 2^2
26y^2 = 4
2y^2 = 4/13
So the areas are 2/5 and 4/13 respectively, and their difference is therefore 6/65. Or if you chose to interpret the question the other way, the first rectangle is 30% larger than the second.
11/18 (0.61111…)
Each term 1/(n(n+3)) can be rewritten as two separate terms:
A/n + B/(n+3)
To find the values of A and B we force them back into a single fraction and make sure it’s equal to the original fraction:
(A(n+3) + Bn)/ (n(n+3)) = 1/(n(n+3))
(A+B)n + 3A = 1
The n term must disappear as there isn’t one on the right-hand side, so A+B=0.
3A=1, so A=1/3, and so B=-1/3
So each term 1/(n(n+3)) is equivalent to 1/3n - 1/(3(n+3)). For example the first term 1/4 is equal to (1/3 - 1/12), the second term 1/10 is equal to (1/6 - 1/15), etc.
This might seem to be unnecessarily complicating the infinite sum, however you might notice that the negative part of each term will appear again as the positive part, three terms later:
(1/3 - 1/12) + (1/6 - 1/15) + (1/9 - 1/18) + (1/12 - 1/21) + (1/15 - 1/24) + (1/18 - 1/27) …
Virtually everything will cancel out. Only the positive parts of the first three terms will remain, and can easily be added together to find the final answer:
1/3 + 1/6 + 1/9 = 11/18, or 0.6111…
If you were to add up successive terms on a spreadsheet you would find that it does indeed approach this sum, but it does so very very slowly. It only reaches 0.6 after about 90 terms, 0.61 after about 900 terms, 0.611 after about 9000 terms, 0.6111 after about 90000 terms etc.
You can use Pythagoras’ theorem to find the length of the diagonal of the block to be 425cm. This diagonal is simultaneously the hypotenuse of another triangle whose base is 340cm and whose height is the number we are after. Using Pythagoras again we find that the height difference is 255cm.
From guess 3, one is in the correct place. It cannot be the initial Red as that would have shown as in the correct place in guess 1. Similarly it cannot be the final purple, as guess 2 would have shown it. So it must be the red in position 2.
Since the first two guesses cover all of the colours, but only show two correct pegs, there must be a colour repeated. It cannot be the red, or else that would have shown in guess 3. It must be one of the colours from guess 2, but since they must occupy positions 1 and 3 in the answer, they cannot be Orange or Purple, as they would have shown as in the correct place in guess 2.
The final answer is therefore Blue Red Blue
All three coins will be at (13,7).
The key to solving this is to compute the x and y coordinates separately. We are given the first three x coordinates, 0, 3, 9 and each subsequent x coordinate will be 3/4 of the previous two less 1/2 of the one before that. Eventually this settles on an x coordinate of 13. Doing the same for the y coordinates will eventually settle on 7.
In general, given starting positions of A, B, C, the coins will converge on: (-2A+B+4C)/3.
The centroid of a shape is the weighted average of the centroids of its component parts.
A_1.C_1 + A_2.C_2 = A.C
Where A_1 is the area of the unshaded region, C_1 is the centroid height of the unshaded region etc.
We require C_2, so making that the subject:
C_2 = (A.C - A_1.C_1)/A_2
We can calculate the area and centroid height (A and C) of the overall figure from first principles, as it is simply an equilateral triangle, and we can find A_2 by simply taking A_1 from A.
C_2 = (rt(3)/4 x rt(3)/6 - rt(3)/6 x rt(3)/10) / (rt(3)/12)
C_2 = 3rt(3)/10 =~ 0.5196…
I'm found in socks, scarves and mittens; and often in the paws of playful kittens. What am I?
Wool!
When the two free endpoints are 0.414213… (sqrt(2) - 1) away from the lower left / upper right corner respectively, the shaded area will be at a maximum, which will be exactly 3.5 - sqrt(8), or 0.67157…, a little over two thirds.
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