Puzzles

Carol will get 270 whatever she does.

The sum of the coefficients can be found by substituting \(x=1\) into the expression. Hence the sum of the coefficients in the expansion of \((2x+1)^5\) is \(3^5\), or 243.

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.