Very few people seem to have attempted this problem, so here is a little help:
5 and 12
16 and 15
3 and 10
they have no common factors.
25 and15 (5 is a common factor)
12 and 16 (2 and 4 are common factors)
3 and 12 (3 is a common factor)
a straight line to (2,1) crosses 2 squares
a straight line to (3,1) crosses 3 squares
a straight line to (x,1) crosses ? squares
a straight line to (p,q) crosses ? squares. Why?
Here is the solution which arrived only just in time to be included!
Dear Cambridge,
We have found a solution to the October six puzzle Beeline.
First we drew up a table showing the number of squares crossed in certain examples
| P | Q | squares crossed |
| 1 | 2 | 2 |
| 1 | 3 | 3 |
| 1 | 4 | 4 etc |
| 2 | 3 | 4 |
| 2 | 5 | 6 |
| 2 | 7 | 8 |
| 3 | 4 | 6 |
| 3 | 5 | 7 |
| 3 | 7 | 9 |
From this we can see that the number of squares crossed is likely to be P+Q - 1
We can prove this by showing that when you cross from (0,0) to (3,4) you are going along 3 squares to the right and up 4 squares. However you cannot count the corner square twice therefore it is 3 squares (P) to the right plus 4 squares (Q) up take away one for the corner square, or P+Q -1.
Happy Hallowe'en love Prav and Sheli age 14 from the North London Collegiate School Maths Puzzle Club.