The coins currently in circulation in Britain are for 1p, 2p, 5p, 10p, 20p, 50p, 100p and 200p. When I gave one of each of these coins to Tom and Harry to share between them, each took two or more coins, with each person's coins' total value equal to a prime number of pence. No coins were left over.

I asked Harry whether his share was a specific number of coins that I mentioned. He said "no".

I then asked Harry whether he had the coin of a denomination I specified. He said "yes".

His two answers allowed me to determine the total of the value that Harry had taken. What was that value?

1) Total of all coins = 388

2) Only 10 pairs of primes add to 388:

(383,5), (359,29), (347,71), (317,71), (281,107), (257,131),

(251,137), (239,149), (197,191)

2) But the following cannot by formed with 2 or more coins:

5, 29, 41, 149, or 191

So Harry's coins must be one of the following:

3 coins:

71 (50+20+1)

107 (100+5+2)

251 (200+50+1)

4 coins:

131 (100+20+10+1)

257 (200+50+5+2)

5 coins:

137 (100+20+10+5+2)

281 (200+50+20+10+1)

317 (200+100+10+5+2)

3) If he does not have 5 coins, and if one of his coins is a 10p, then
he must have 131p. This is the only pair of answers which would
allow a determination of his coins. QED, this is the case.

*Edited on ***September 17, 2005, 3:55 pm**