Puzzles

There are 1 time starting 00; 2 times starting 01; 3 times starting 02; 4 times starting 03; 5 times starting 04; 6 times starting 05; 6 times starting 06; 6 times starting 07; 6 times starting 08; 6 times starting 09; 2 times starting 10; 3 times starting 11; 4 times starting 12; 5 times starting 13; 6 times starting 14; 6 times starting 15; 6 times starting 16; 6 times starting 17; 6 times starting 18; 5 times starting 19; 3 times starting 20; 4 times starting 21; 5 times starting 22; and 6 times starting 23.

Therefore there are 112 ways in total.

The product of the red digits is 192.

Carol will get 270 whatever she does.

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.

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.