Parallelogram 15 • Level 2 • 14 Dec 2023Chrismaths - Part 2

There are 60 − 22 = 38 minutes from 22:22 to 23:00 and then a further 60 minutes to midnight.

Since 38 + 60 = 98, there are 98 minutes to midnight.

In the context of the JMC we are entitled to assume the truth of the statement that 123456789×8 is obtained by interchanging two of the digits of 987654321. This leads to a quick way to answer the question without the need for a lot of arithmetic.

Because 9×8=72, the units digit of 123456789×8 is a 2. Starting from 987654321, to obtain a 2 as the units digit we need to interchange the digits 1 and 2. So these are the two digits which are in a different order in the answer to the calculation. Now comes the easy bit, 1+2=3.

Alternatively, if we cannot take the statement in the question on trust, the only thing to do is to actually multiply 123456789 by 8. If you do this you will see that the answer is 987654312. It follows that it is the digits 1 and 2 that need to be interchanged.

Note that, in fact, as soon as we get as far as working out that 89×8=712 we can deduce that the digits 1 and 2 need to be interchanged. It is, however, necessary to do the whole sum to check that all the other digits are in the right order.

Note The number 123456789×8 is a multiple of 8. We have the following test for whether a number is a multiple of 8.

An integer is a multiple of 8, if, and only if, its last three digits form a number which is a multiple of 8.

Since 321 is not a multiple of 8, this shows immediately that 987654321 is not equal to 123456789×8 .

Since the fish weighs 2 kg plus one third of its weight, 2 kg is two thirds of its weight. Therefore one third of its weight is 1 kg, and so the total weight of the fish is 2 kg + 1 kg = 3 kg.

Alternatively, we can also solve this problem using algebra.

We let x be the weight of the fish in kg. Now we use the information in the question to create an equation involving x that we can solve.

Because the fish weighs 2 kg plus one third of its weight, x=2+13x

It follows that x−13x=2, and hence, 23x=2.

Because 32×23=1, we multiply both sides of this equation by 32. In this way we deduce that x=32×2=3.

Either the Knave of Hearts stole the tarts or he is innocent.

If the Knave of Hearts stole the tarts, he was telling the truth. So the Knave of Clubs was lying. Hence the Knave of Diamonds was telling the truth. Therefore the Knave of Spades was lying. So in this case two of the four Knaves were lying.

If the Knave of Hearts did not steal the tarts, he was lying. So the Knave of Clubs was telling the truth. Hence the Knave of Diamonds was lying. Therefore the Knave of Spades was telling the truth. So also in this case two of the four Knaves were lying.

We cannot tell from the information given whether or not the Knave of Hearts stole the tarts. But, as we have seen, we can be sure that, whether he stole them or not, two of the Knaves were telling the truth and two were lying.

We form a complete grid inside the hexagon, as shown in the figure.

In this way the hexagon is divided up into a number of congruent equilateral triangles and, around the edge, some triangles each congruent to half of the equilateral triangles.

We could now use the grid to work out the shaded and unshaded areas in terms of the areas of the equilateral triangles, and hence work out which fraction of the area of the hexagon is shaded.

It is a little easier to exploit the sixfold symmetry of the figure and just work out the fraction of the area surrounded by the heavy lines that is shaded.

We see that in this part of the hexagon there are six shaded equilateral triangles, four unshaded equilateral triangles and four unshaded triangles whose areas are each half that of the equilateral triangles. So the unshaded area is equal to that of six of the equilateral triangles. It follows that the shaded area is equal to the unshaded area.

We conclude that the fraction of the hexagon that is shaded is 12.