Mathematical & Logical Puzzles -A5

17.     Missionaries and Cannibals

M = misionary      C = cannibal         Bracketed letters are the ones who can row.

Left bank	River	Right bank
MM(M)CC(C)	-	-
MMC(C)	(M)C ®	-
MMC(C)	¬ (M)	C
MM(M)	(C)C ®	C
MM(M)	¬ (C)	CC
M(C)	(M)M ®	CC
M(C)	¬ (M)C	MC
MC	(M)(C) ®	MC
MC	¬ (M)C	M(C)
CC	(M)M ®	M(C)
CC	¬ (C)	MM(M)
C	(C)C ®	MM(M)
C	¬ (C)	MM(M)C
-	(C)C ®	MM(M)C
-	-	MM(M)CC(C)

A total of thirteen crossings.
18.     Road Safety Addition

There are seven basic solutions, values for D and H are the two remaining digits in each case.  These two digits can be put in either order making 14 solutions altogether.  If zero cannot be the first digit, #4 has only one, not two solutions (H not equal 0) making only 13 solutions in total.

	R	T	A	M	L	O	J	E
1	2	9	7	6	1	0	5	3
2	2	9	7	6	0	1	4	3
3	2	9	7	6	0	1	5	4
4	2	9	6	5	4	8	7	3
5	2	4	7	6	3	9	0	8
6	6	7	3	2	0	5	9	8
7	6	7	3	2	4	1	5	0

19.     The Lonesome Eight

When two digits are brought down instead of one, there must be a zero in the quotient.  This occurs twice, so we know at once that the quotient must be x080x.

When the divisor is multiplied by the quotient’s last digit, the produxt is a four-digit number.  But we see that 8 times the divisor is a three-digit number.  So the quotient’s last digit must be nine.

The divisor must be less than 125 because 8 ´ 125 = 1000, a four-digit number. 

We can now deduce that the quotient’s first digit must be more than 7, for 7 times a divisor less than 125 would give a product which would be more than 100 less than 1000, the lowest four-digit number, and therefore would give a difference of more than 100 (a three digit number) when subtracted from the first four digits of the dividend, instead of the two-digit difference it does, in fact, give.

The first digit of the quotient cannot be nine because nine times the divisor is a four figure number, so it must be 8.  This makes the full quotient 80809.

The divisor must be more than 123 because 123 ´ 80809 is a seven-digit number and our dividend has eight digits.  Therefore it must be 124.

The final result is now easily found.

20.     The Farmer And The Hundred Dollars

To buy 100 animals with $100, the number of anomals bought must equal the number of dollars spent.  Since sheep sell at the price of one sheep for $1, they can be ignored until we have bought a number of cows and pigs such that the number of cows and pigs bought, equals the number of dollars paid for them.  The rest of the $100 can be made up of sheep, which will not affect the balance of animals bought against dollars spent.

For every cow bought, we have spent $4 more than we need to, to make the number of cows bought equal to dollars spent.  For every pig bought, we have spent 95 cents less than necessary to make the number of pigs bought equal to the number of dollars spent.

Thus if we buy cows and pigs in the inverse ratio of this excess and insufficiency, (i.e. 19 cows for every 80 pigs) the number of animals bought will be equal to the number of dollars spent.

But 80 + 19 = 99 so we can only buy one such balanced set for under $100.  The other $1 can be used to buy a sheep.  Therefore the farmer bought 19 cows, 1 sheep and 80 pigs.


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