Expressing 51 and 41 with a common denominator of 40, we get 51=408 and 41=4010. We require that 40n>408 and 40n<4010, thus n>8 and n<10. The only integer n that satisfies both of these inequalities is n=9. Expressing 8m and 31 with a common denominator of 24, we require 243m>248 and so 3m>8 or m>38. Since 38=232 and m is an integer, then m≥3. Expressing 8m+1 and 32 with a common denominator of 24, we require 243(m+1)<2416 or 3m+3<16 or 3m<13, and so m<313. Since 313=431 and m is an integer, then m≤4. The integer values of m which satisfy m≥3 and m≤4 are m=3 and m=4. At the start of the weekend, Fiona has played 30 games and has w wins, so her win ratio is 30w. Fiona’s win ratio at the start of the weekend is greater than 0.5=21, and so 30w>21. Since 21=3015, then we get 30w>3015, and so w>15. During the weekend Fiona plays five games giving her a total of 30+5=35 games played. Since she wins three of these games, she now has w+3 wins, and so her win ratio is 35w+3. Fiona’s win ratio at the end of the weekend is less than 0.7=107, and so 35w+3<107. Rewriting this inequality with a common denominator of 70, we get 702(w+3)<7049 or 2(w+3)<49 or 2w+6<49 or 2w<43, and so w<243. Since 243=2121 and w is an integer, then w≤21. The integer values of w which satisfy w>15 and w≤21 are w=16,17,18,19,20,21.


