 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Divide into Primes (Posted on 2005-09-17) 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?

 See The Solution Submitted by Charlie Rating: 3.2857 (7 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 12
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
 Posted by Steve Herman on 2005-09-17 03:33:17 Please log in:

 Search: Search body:
Forums (0)