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Postman delivers the mail in the small village. This village has exactly ten houses, numbered from 1 to 10.

In a certain week, he did not deliver any mail at two houses in the village but at the other houses he delivered mail three times each.

Each working day he delivered mail at exactly four houses.

The sums of the house numbers where he delivered mail were:

on Monday: 18

on Tuesday: 12

on Wednesday: 23

on Thursday: 19

on Friday: 32

on Saturday: 25

on Sunday: he never works

Which two houses didn't get any mail that week?

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  • $\begingroup$ This puzzle would've benefited from a no-computers tag... $\endgroup$ – EKons Aug 28 '18 at 12:21
  • $\begingroup$ Not recommended to change the tags now that an answer has been accepted. I was just underway making a program to calculate this, integer partitions. :P $\endgroup$ – EKons Aug 28 '18 at 12:30
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    $\begingroup$ @ΈρικΚωνσταντόπουλος The tag is welcome. Even if an answer was accepted, it's still interesting to know that this puzzle was "designed to be solved without using calculators or computer programming". $\endgroup$ – xhienne Aug 28 '18 at 12:36
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I think the house numbers which don't get mail are

$4$ and $8$

Reasoning

Given the premise of the problem, the sum of the numbers over the whole week should be equal to three times the sum of the numbers of the houses that get mail, that is

$ 18 + 12 + 23 + 19 + 32 + 25 = 129 = 3 \times 43$ (Thank you, Marius!)

And so the sum of the numbers of the houses which don't get mail is $55 - 43 = 12$.

Hence, the only possibilities for houses which don't get mail are the pairs $(5,7)$, $(4,8)$, $(3,9)$ and $(2,10)$.

Given that on Friday, the sum of the four houses that get mail is $32$, which is greater than $9 + 8 + 7 + 6$ and $10 + 8 + 7 + 6$, this means that both houses $9$ and $10$ must receive mail.

Also $32$ cannot be made with four houses without using either $5$ or $7$ (since $10+9+8+6 > 32$ and $10+9+8+4 < 32$) so the only remaining possibility is $(4,8)$ for the houses that don't get mail.

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    $\begingroup$ love the reasoning, but you should use a calculator next time...do the math again :) specially when dividing something by 3 (or multiplying...depends on how you see it) $\endgroup$ – Marius Aug 28 '18 at 12:04
  • $\begingroup$ @Marius Thank you, completely missed that. $\endgroup$ – hexomino Aug 28 '18 at 12:08
  • $\begingroup$ You're welcome. Just noticed because I started on the same approach and got a different sum $\endgroup$ – Marius Aug 28 '18 at 12:08
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    $\begingroup$ I was too slow on this one. I went for the small numbers instead and used Tuesday's 12 to eliminate most options but was still stuck on the last two possibilities when I saw your answer. $\endgroup$ – Jaap Scherphuis Aug 28 '18 at 12:18
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Tuesday's delivery would have been 1+2+3+6 or 1+2+4+5. Friday's delivery would have been 5+8+9+10 or 6+7+9+10. This means the solution is a pair picked from one of the 4 combinations 4+7, 4+5+8, 3+6+7, 3+8. The total delivered house numbers were 129. Had every house gotten 3 deliveries, it would have been 55*3=165, so the sum of the missing numbers is (165-129)/3 = 12. Only 4+8 fits the bill.

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10, 9,7,6, 5,3,2,1 get mail, and 8 and 4 dont. 10 and 9 should be in the 32 combined with either 8 and 5 or 7 and 6. The total sum is 129, knowing that 57 (for 10 and 9) and 39 (either 8 and 5 or 7 and 6) could be substracted. Leaving 33 which is relative low, so i calculated (1 +2 +3 +4 ) *3 =30, and followed that the 4 needs to be replaced by the 5.

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