Earlier today I set you these five puzzles from Mathigon’s advent calendar. Here they are again, with solutions. Sorry if you came here to read about Ian Dury, but in recompense you get to get your head around these blocks:

**1. Hit me with your four cube stick**

Every cube needs to touch at least one other cube. Faces have to line up. Ignore rotations and reflections. Don’t rush this one! It is very easy to miss obvious constructions.

**Solution**

Photograph: Mathigon

**2. Timelines**

**Solution **22 times in 24 hours

If we consider 6:00 AM as the first time, the hands will form a straight line every 1 hour, 5 minutes, and 27 seconds.

**3. Eight heavy adult females**

**Solution **888+88+8+8+8 = 1000

I also asked about how to make 100. That’s 88 + 8 + ((8 + 8 + 8 + 8)/8)

Or 10. That’s 8 + ((8 + 8)/8) + 8 + 8 – 8 – 8

**4. Slice of Orange**

A familiarity with the Pythagorean theorem may help. It says that for right-angled triangles, the square of the hypotenuse is equal to the squares of the other two sides

**Solution **π/6

solution-1

Let C be the centre of the semicircle, and let r be the radius of the semicircle. We can make the right-angled triangle (shaded yellow) OCB that has two sides of length r. Using Pythagoras, the hypotenuse of this triangle has length r√2. Now we use Pythagoras on the red triangle OCE, which has hypotenuse of 1, since this is simple a radius of the large circle. If (r√2)2 + r2 = 12 then 3r2 = 1, or r = 1/√3.

The area of a circle of radius r is πr2, so the area of a semicircle of radius r is πr2/2.

The answer is π(1/√3)2/2 = π/(3x2) = π/6

**5. Bucket list**

**Solution **1200/3125, or 38.4per cent

There are 5⁵ = 3125 total arrangements possible (five balls in five buckets).

The only way to get a single empty bucket is if one other bucket has 2 balls, and the others have 1 ball.

There are 5 ways to pick an empty bucket.

Next, there are 4 ways to pick the bucket with two balls.

There are “5-choose-2” = 10 ways to pick the two balls that get doubled up in the same bucket.

There are 3x 2 = 6 ways to distribute the remaining 3 balls in the remaining 3 buckets.

Thus there are 5×4×10×6 = 1200 arrangements with exactly one bucket empty.

The probability is 1200/3125 = 38.4%

UPDATE: Originally there was a solution here saying that the answer was 120/286, or about 42%. As it has been pointed out, this answer was wrong. The correct answer is the one above.

Thanks to Mathigon for today’s puzzles, which appear in their forthcoming Mathigon advent calendar. Mathigon is a free website about maths, with fantastic interactive features. Highly recommended!

I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.

I give school talks about maths and puzzles (online and in person). If your school is interested please get in touch.

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