Postman Nat delivers the mail in a small village which has only one street with exactly ten houses, numbered from 1 up to and including 10.
In a certain week, Nat did not deliver any mail at two houses in the village; 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
op Saturday: 25
on Sunday: he never works
Which two houses didn't get any mail that week?
Houses 4 and 8 received no mail that week.
The first line in each pair of lines below is the same, indicating the number of mail deliveries in houses 1 to 10 in order. The second of the two lines is the day-by-day itinerary of the postman; there are 13 possible arrangements of these.
3 3 3 0 3 3 3 0 3 3
1 2 5 10 | 1 2 3 6 | 1 5 7 10 | 2 3 5 9 | 6 7 9 10 | 3 6 7 9
3 3 3 0 3 3 3 0 3 3
1 2 5 10 | 1 2 3 6 | 1 6 7 9 | 2 3 5 9 | 6 7 9 10 | 3 5 7 10
3 3 3 0 3 3 3 0 3 3
1 2 5 10 | 1 2 3 6 | 2 5 7 9 | 1 3 5 10 | 6 7 9 10 | 3 6 7 9
3 3 3 0 3 3 3 0 3 3
1 2 5 10 | 1 2 3 6 | 2 5 7 9 | 1 3 6 9 | 6 7 9 10 | 3 5 7 10
3 3 3 0 3 3 3 0 3 3
1 2 5 10 | 1 2 3 6 | 3 5 6 9 | 1 2 7 9 | 6 7 9 10 | 3 5 7 10
3 3 3 0 3 3 3 0 3 3
1 2 6 9 | 1 2 3 6 | 1 5 7 10 | 2 3 5 9 | 6 7 9 10 | 3 5 7 10
3 3 3 0 3 3 3 0 3 3
1 2 6 9 | 1 2 3 6 | 2 5 7 9 | 1 3 5 10 | 6 7 9 10 | 3 5 7 10
3 3 3 0 3 3 3 0 3 3
1 3 5 9 | 1 2 3 6 | 1 5 7 10 | 2 3 5 9 | 6 7 9 10 | 2 6 7 10
3 3 3 0 3 3 3 0 3 3
1 3 5 9 | 1 2 3 6 | 2 5 6 10 | 1 2 7 9 | 6 7 9 10 | 3 5 7 10
3 3 3 0 3 3 3 0 3 3
1 3 5 9 | 1 2 3 6 | 2 5 7 9 | 1 2 6 10 | 6 7 9 10 | 3 5 7 10
3 3 3 0 3 3 3 0 3 3
1 3 5 9 | 1 2 3 6 | 2 5 7 9 | 1 3 5 10 | 6 7 9 10 | 2 6 7 10
3 3 3 0 3 3 3 0 3 3
2 3 6 7 | 1 2 3 6 | 1 5 7 10 | 2 3 5 9 | 6 7 9 10 | 1 5 9 10
3 3 3 0 3 3 3 0 3 3
2 3 6 7 | 1 2 3 6 | 2 5 7 9 | 1 3 5 10 | 6 7 9 10 | 1 5 9 10
The program assumes the two house numbers for non-delivery add up to 12 so that the total of Monday's through Saturday's deliveries adds up to 36 less than three times the sum of the numbers 1 through 10.
DECLARE SUB deliver (day!)
DIM SHARED nd1, nd2, dels(10), hist(30, 4), solct
CLS
FOR nd1 = 2 TO 5
nd2 = 12 - nd1
deliver 1
NEXT
SUB deliver (day)
FOR h1 = 1 TO 7
IF h1 <> nd1 AND h1 <> nd2 AND dels(h1) < 3 THEN
FOR h2 = h1 + 1 TO 8
IF h2 <> nd1 AND h2 <> nd2 AND dels(h2) < 3 THEN
FOR h3 = h2 + 1 TO 9
IF h3 <> nd1 AND h3 <> nd2 AND dels(h3) < 3 THEN
FOR h4 = h3 + 1 TO 10
IF h4 <> nd1 AND h4 <> nd2 AND dels(h4) < 3 THEN
tot = h1 + h2 + h3 + h4
good = 0
SELECT CASE day
CASE 1: IF tot = 18 THEN good = 1
CASE 2: IF tot = 12 THEN good = 1
CASE 3: IF tot = 23 THEN good = 1
CASE 4: IF tot = 19 THEN good = 1
CASE 5: IF tot = 32 THEN good = 1
CASE 6: IF tot = 25 THEN good = 1
END SELECT
IF good THEN
dels(h1) = dels(h1) + 1
dels(h2) = dels(h2) + 1
dels(h3) = dels(h3) + 1
dels(h4) = dels(h4) + 1
hist(day, 1) = h1
hist(day, 2) = h2
hist(day, 3) = h3
hist(day, 4) = h4
IF day = 6 THEN
FOR i = 1 TO 10
PRINT dels(i);
NEXT
PRINT
FOR d = 1 TO 6
FOR i = 1 TO 4
PRINT USING " ##"; hist(d, i);
NEXT: IF d < 6 THEN PRINT "|";
NEXT: PRINT : PRINT
ELSE
deliver day + 1
END IF
dels(h1) = dels(h1) - 1
dels(h2) = dels(h2) - 1
dels(h3) = dels(h3) - 1
dels(h4) = dels(h4) - 1
END IF
END IF
NEXT h4
END IF
NEXT h3
END IF
NEXT h2
END IF
NEXT h1
END SUB
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Posted by Charlie
on 2013-06-11 17:24:10 |
computer solution
