The Wandering Astronomer

Let’s figure out which star goes into which box.

Clue # 1: The volume of the Canopus box is greater than the combined volumes of the Altair and Sirius boxes. Sirius’ box is larger than Altair’s.
At this point, we can only conclude that Canopus is larger than both Altair and Sirius, We also know that Sirius’s box is larger than Altair’s.

Clue # 2: The side length of the box holding one Orion star is one meter greater than the side length of the box holding the other Orion star.
The two Orion stars are Betelgeuse and Rigel. We only know that the the length of one of their boxes is one inch greater than the length of the other.

Clue # 3: The numerical value of the volume of Beta Centauri’s box is equal to the numerical value of its surface area.
The volume of a box equals the cube of its side length. The surface area of a box equals 6 tiimes the square of its side length. We don’t need to scroll through the volumes and surface areas of all boxes. Insteasd, we can set s^3 = 6s^2.
Dividing both sides by s^2 reduces the equation to s = 6.
Beta Centauri, also known as Hadar, must be in the box with a side lenfth of 6 meters.

1 –
2 –
3 –
4 –
5 –
6 – Hadar
7 –
8 –
9 –
10 –
11 –
12 –

Clue # 4: The side length of Acrux’s box is equal to the average side lengths of the boxes containing Sirius and Canopus
We know that Acrux’s box must be between the boxes containing Sirius and Canopus,

Clue # 5: The numerical value of the volume is equal to the numerical value of the side length of the box containing one of the winter triangle stars.
The only box for which this statement is true is the one with a 1 inch side length. 1^3 = 1.
All three of the Winter Triangle Stars (Procyon, Betelgeuse and Sirius) appear on this list.
Because of Clue # 1, we know that Sirius cannot be in this box as its box has to be larger than Altair’s.
It must either be Procyon or Betelgeuse.

Clue # 6: The numerical value of the surface area of the box containing the 3rd brightest star on the list is equal to the atomic number of chromium.
The atomic number of chromium is 24.
The surface area of a box equals 6s^2
6s^2 = 24
s^2 = 4
s = 2
Arcturus is the 3rd brightest star.
It belongs in rhe second box.

1 –
2 – Arcturus
3 –
4 –
5 –
6 – Hadar
7 –
8 –
9 –
10 –
11 –
12 –
We can now determine that Procyon must be in the first box. Remember that Clue # 2 states that the box containing one Orion star must be one inch greater than the box containing the other.
Since the second box doesn’t contain Rigel, the first box cannot contain Betelgeuse.

1 – Procyon
2 – Arcturus
3 –
4 –
5 –
6 – Hadar
7 –
8 –
9 –
10 –
11 –
12 –

Clue # 7: The ratio of the numerical value of the volume of Canopus’ box and that of the box’s surface area is 10:6
The volume of the 10th box is 1000. The surface area is 600. This reduces to a ratio of 10:6.
Canopus is in the 10th box,

1 – Procyon
2 – Arcturus
3 –
4 –
5 –
6 – Hadar
7 –
8 –
9 –
10 – Canopus
11 –
12 –

Clue # 8: If the volume of Betelgeuse’s box is 2^n, the volume of the box containing Sirius is 2^(n+3)
Let’s regard the box volumes:

side length volume
1 1
2 8
3 27
4 64
5 125
6 216
7 343
8 512
9 729
10 1000
11 1331
12 1728

Let’s now regard the powers of 2

n 2^n
0 1
1 2
2 4
3 8
4 16
5 32
6 64
7 128
8 256
9 512
10 1024
11 2048
12 4096

The only volumes that are also equal to powers of 2 are 1 (box 1) 8 (box 2) , 64 (box 4) , and 512 (box 8).
Procyon is already in box 1, Arcturus is already in box 2. That leaves only boxes 4 and 8,
If the volume of box 4 is 2^n, the volume of box 8 will equal 2^(n+3).
Betelgeuse is in box 4.
Sirius is in box 8.

1 – Procyon
2 – Arcturus
3 –
4 – Betelgeuse
5 –
6 – Hadar
7 –
8 – Sirius
9 –
10 – Canopus
11 –
12 –
Returning to Clue # 4: The side length of Acrux’s box is equal to the average side lengths of the boxes containing Sirius and Canopus
The sum of the side lengths of Sirius and Canopus equals 18.
The average of these values is 9. Acrux must be in the 9th box.

1 – Procyon
2 – Arcturus
3 –
4 – Betelgeuse
5 –
6 – Hadar
7 –
8 – Sirius
9 – Acrux
10 – Canopus
11 –
12 –
‘
Clue # 9: The numerical value of the surface area of Capella’s box is a factor of 450
The factors of 450: 1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90, 150, 225, and 450.
The only surface area values that are factors of 450 are:
Box 1: 6
Box 5: 150
Procyon is already in Box 1.
Capella must be in Box 5.

1 – Procyon
2 – Arcturus
3 –
4 – Betelgeuse
5 – Capella
6 – Hadar
7 –
8 – Sirius
9 – Acrux
10 – Canopus
11 –
12 –
Since Capella is in box 5, Rigel must be in box 3. Remember that the side length of the box containing one Orion star is only one greater than the box containing the other.

1 – Procyon
2 – Arcturus
3 – Rigel
4 – Betelgeuse
5 – Capella
6 – Hadar
7 –
8 – Sirius
9 – Acrux
10 – Canopus
11 –
12 –
Returning to Clue # 1: The volume of the Canopus box is greater than the combined volumes of the Altair and Sirius boxes. Sirius’ box is larger than Altair’s.
Altair cannot be larger than Canopus, so it must be in box # 7.

1 – Procyon
2 – Arcturus
3 – Rigel
4 – Betelgeuse
5 – Capella
6 – Hadar
7 – Canopus
8 – Sirius
9 – Acrux
10 – Canopus
11 –
12 –

Clue # 10. The last letter of the name of the star contained within the largest box is the same as the first letter of the name of the star in the second to largest box.
We have only two stars remaining. Working out their positions requires alphabetical knowledge rather than mathematical reasoning.
Vega and Achernar.
Vega must be in box # 12 and Achernar must be in box # 11.

1 – Procyon
2 – Arcturus
3 – Rigel
4 – Betelgeuse
5 – Capella
6 – Hadar
7 – Canopus
8 – Sirius
9 – Acrux
10 – Canopus
11 – Achernar
12 – Vega