The Ultimate Puzzle Site - Puzzle Solution

Solution to: Postman Pat

If postman Pat would have delivered mail three times at each house, then the total sum of the house numbers per day would be (1+2+3+4+5+6+7+8+9+10)×3=165. Now that sum is 18+12+23+19+32+25=129. The difference is 165-129=36; divided by 3 this is 12. The sum of the house numbers where no mail was delivered is therefore 12. The following combinations are possible:

2+10
3+9
4+8
5+7

Each day at four houses the mail was delivered. On Tuesday the sum was 12. 12 can only be made from four house numbers in 2 ways:

1+2+3+6
1+2+4+5

The same holds for Friday with the sum of 32:

5+8+9+10
6+7+9+10

From this we can conclude that the house numbers 1, 2, 9 and 10 for sure have received mail, which means that the combinations 2+10 and 3+9 are not possible. Also the combination 5+7 is not possible, because mail was delivered either at house 5 or at house 7. Thus the only remaining solution is: houses 4 and 8.

N.B.: there are various possibilities for the actual post delivery of the whole week. For example:

Monday	houses 1, 3, 5 and 9
Tuesday	houses 1, 2, 3 and 6
Wednesday	houses 1, 5, 7 and 10
Thursday	houses 2, 3, 5 and 9
Friday	houses 6, 7, 9 and 10
Saturday	houses 2, 6, 7 and 10