Puzzles

The only way to make 208 by adding consecutive numbers is 10+11+12+13+14+15+16+17+18+19+20+21+22. Therefore next year Noel will give his grandchildren a total of £221.

Pick one of the 28 points and imagine a line to the point opposite it. There will be 13 points on each side of this line. Picking one of these 13 points and then picking the corresponding point on the other side of the line gives an isosceles triangle. Therefore there are 13 isosceles triangles with the first chosen point as the point where the two equal sides meet.

There were 28 choices for the first point, and so the total number of isosceles triangles will be 13×28=364.

Let \(a_n\) be the number of ways to tile an \(n\times2\) rectangle.

There is one way to tile a 1×2 rectangle; and there are three ways to tile a 2×2 rectangle. Therefore \(a_1=1\) and \(a_2=3\).

In a \(n\times2\) rectangle, the rightmost column will either contain a vertical 2×1 domino, two horizontal 2×1 dominoes, or a 2×2 square. Therefore \(a_n=a_{n-1}+2a_{n-2}\).

Using this, we find that there are 341 ways to tile a 9×2 rectangle.

The largest number you can make with the digits in the red boxes is 432.

In order to make a valid triangle, the sum of the two shorter sides must be larger than the longest side. There is 1 triangle with longest side 4cm (2,3,4); 2 with longest side 5cm (2,4,5 and 3,4,5); 4 with longest side 6cm (2,5,6; 3,5,6; 4,5,6 and 3,4,6); 6 with longest side 7cm (2,6,7; 3,6,7; 4,6,7; 5,6,7; 3,5,7 and 4,5,7); 9 with longest side 8cm; 12 with longest side 9cm; 16 with longest side 10cm; 20 with longest side 11cm; 25 with longest side 12cm; 30 with longest side 13cm; 36 with longest side 14cm; and 42 with longest side 15cm.

In total this makes 203 triangles.

100 numbers will have 1 in the units column. Another 100 with have a 1 in the tens column; and other 100 will have a 1 in the hundreds column. Finally, 1000 has a 1 in the thousands column.

In total, this makes 301 copies of the digit 1.

I will use R, B and G to represent the heavy weights and r, b and g to represent the lighter weights. There are a few ways of doing this, the first way I found was the following method.

First, put one red and one blue on both ends of the scale. Either (a) you have both heavy weights on one side, so have RB against rb and the RB side is heavier; or (b) there is one heavy weight on each side, so you have Rb against rB and the scales will balance.

If (a), you will know which red and blue weights are the heaviest. You can then weigh the green weights against each other and you have found all three heavy weights.

If (b), then you know that the red on one side of the scale is equal to the blue on the other side. You can then put one red and one green on both ends of the scale, and apply the same logic to work out which blues and reds are equal, and you are done.

Interestingly, if the scales balance on both weighings, then you will know which weights are equal to each other, but you will not know which three are the heavier.

You have \(2n\) weights. Two of them are colour 1, two are colour 2, ..., two are colour \(n\). One weight of each colour is heavier than the other; the \(n\) heavy weights all weigh the same, and the \(n\) lighter weights also weigh the same.

How many times do you need to use the sales to split the weights into two sets by weight?

For any digit \(V\), \(VVV\) is a multiple of 111, and \(111=37\times3\). 37 is prime, so \(VI\) must be a multiple of 37 (as \(X\) is less than 10 so cannot be a multiple of 37). Therefore \(VI\) is either 37 or 74.

If \(VI\) was 74, then \(VVV\) is 777. But then \(VI\times X\) is even and \(VVV\) is odd. This is impossible, so \(VI\) must the 37.

Now that we know that \(VI\) is 37, we see that

and so \(X=9\) and the final solution is