Puzzles

The directions on each die except the last must be followed. Therefore (for example) blue cannot be up, as at least one of the blue dice in the top row must be used. This would lead to the route leaving the square.

By a process of elimination as above, red is down, blue is right, orange is up, green is left. This means that the last die visited will be the orange 2 in the middle of the left most column.

a) Always

b) When \(n<\frac{1}{f_x}\), where \(f_x\) is the fractional part of \(x\).

c) When the fractional parts of \(x\) and \(y\) add up to less than one.

d) Let \(f_x\) and \(f_y\) be the fractional parts of \(x\) and \(y\) (respectively).

This will be equal to \(\lfloor x\rfloor \lfloor y\rfloor \) when \(\lfloor f_x\lfloor y\rfloor +f_y\lfloor x\rfloor +f_yf_x\rfloor =0\).

For this to be true, it is necessary (but not sufficient) that \(f_y<\frac{1}{x}\) and \(f_x<\frac{1}{y}\).

Show that

Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums reading across and down are correct.

Let \(p(n)\) denote the number of people in the correct seat when the table is rotated by \(n\) people. From the question, we know that:

As every guest can be put into the correct place by one rotation, we know that:

As \(p(0)=0\):

This sum has 23 positive integers adding up to 24, so one of the rotations must lead to at least two people being in the correct places.

If the 24 guests sat randomly and one person was in the correct seat, could you still rotate the table so that two people are correctly seated?

If \(n\) is odd, the vertices of a regular \(n\)-gon are balanced.

If \(n\) is even, a balanced set can be constructed as follows:

If \(n=4\), this set is balanced (both the triangles are equilateral):

Draw a triangle with its centre at one of the points, going through the other three points:

For \(n>4\), repeatedly add a two new points on the circle which form an equilateral triangle with the centre. For example, for \(n=6\), this set is balanced:

And for \(n=8\), this set is balanced:

This is true for any base larger than 2. This is because:

Which other base 10 square numbers are always square in high enough bases?

The second hand will always be pointing at one of the 60 graduations. If the minute and hour hand are 120° away from the second hand they must also be pointing at one of the graduations. The minute hand will only be pointing at a graduation at zero seconds past the minute, so the second hand must be pointing at 0. Therefore the hand are either pointing at: hour: 4, minute: 8, second: 0; or hour: 8, minute: 4, second: 0. Neither of these are real times, so it is not possible

If the second hand moves continuously instead of moving every second, will there be a time when the hands of the clock are all 120° apart?