Solution to:
Postman Pat
If postman Pat had 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 165129=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 two 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.
In addition, 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 postdelivery 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 
Copyright © 2002 E.R. van Veldhoven. All rights reserved.
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