The coins add up to 388 pence. The possible values that add up to 388 pence, and the divisions of the denominations, are:
5 383 5 vs 1,2,10,20,50,100,200
29 359 no way
41 347 no way
71 317 50,20,1 vs 2,5,10,100,200
107 281 100,5,2 vs 1,10,20,50,200
131 257 100,20,10,1 vs 2,5,50,200
137 251 100,20,10,5,2 vs 1,50,200
149 239 no way
191 197 no way
Since each of the two took more than one coin, 5+383 is out.
Some values split between 3 coins and 5 coins, while 131+257 splits into four coins on each side.
If I had asked if the number of coins was 4, and received that negative answer, then no matter what denomination was confirmed, there'd be more than one possible value that Harry could have.
Likewise if I had asked if the number of coins was 3, then no matter what denomination was confirmed, there'd be more than one possible total among the 4 and 5 coin sets.
If you look at the numbers, you'll see that the 10p denomination is never part of a group of three. So if I had asked if the number of coins was 5, and received a no answer, and then asked if he had taken the 10p piece and the answer was yes, it would have to be the 4-coin set that contains the 10p piece, and that has a value of **131 pence**.
This is **New Scientist**'s Enigma # 1346 from the issue of 25 June 2005. |