A biscuit for the dog

Wacky · 2003-07-03, 17:38

Four tramps bought, borrowed, found, or otherwise came into possession of a tin of biscuits. They agreed that they would evenly divide them in the morning.

During the night, while the others were asleep, one of them opened the tin and found that if he divided the lot into four equal shares, there would be one left over. To eliminate the uneven division, he fed one to the dog and removed exactly 1/4 of the remainder.

Later, a second man did the same. The dog again received a bribe of one biscuit to remain quiet and the man removed exactly 1/4 of the remaining ones.

Subsequently, the third and fourth did likewise.

In the morning, everyone noticed the reduced number of biscuits in the tin, but said nothing since he had removed some of them. So they divided them equally and found one left over, which again went to the dog.

In total, the dog got 5.

What is the minimum number of biscuits that were originally in the tin?

Orgasmic Troll · 2003-07-03, 18:55

85 .. yes?

Jwb52z · 2003-07-03, 19:12

If I understand this, we're not talking about American biscuits, right?

Maybeso · 2003-07-03, 19:16

There were 1021 biscuits in the tin. That's an impressive sized tin.

Wacky · 2003-07-03, 19:26

Quote:

Originally Posted by Jwb52z

If I understand this, we're not talking about American biscuits, right?

Spot on.

Quote:

Originally Posted by Maybeso

That's an impressive sized tin.

Or very small biscuits. :)

Nick was the first to get it right

Quote:

Originally Posted by NickGlover

The answer is 1021.

Bonus question: Can you express the number for N tramps?

Maybeso · 2003-07-03, 20:05

For N tramps, the ___ (N+1)
tin should contain __N___ - (N - 1) biscuits.
[edit]
For N tramps, the tin should contain N^(N+1) - (N - 1) biscuits.
[/edit]
An interesting observation, if the 4 tramps would rather not give the dog any biscuits,
there would need to be 1024 biscuits in the tin -- 3 more, not less.

NickGlover · 2003-07-03, 20:36

Well, I can't read Maybeso's answer (unsupported characters?), but my answer is 4^(N+1) - 3.

Maybeso · 2003-07-03, 21:27

Sorry about that, I tried to get fancy.

For N tramps, the tin should contain N^(N+1) - (N - 1) biscuits.

Nick, you have the 4 and the 3 as a constant, but I think they also have to vary with N.
(It is nice to see we're thinking along the same lines.) ;)

NickGlover · 2003-07-03, 21:54

Oh, you are right that my answer is wrong in keeping the 4 and 3. Intuitively, your answer seems right, but I don't have time to verify it right now. I will look at it later this evening if Wackerbarth hasn't confirmed it by then.

Wacky · 2003-07-03, 22:23

For N tramps, the tin should contain N^(N+1) - (N - 1) biscuits.

Spot On, Gentlemen!

I'll have to avoid such easy ones ....

koal · 2003-07-04, 15:20

Had a similar idea years ago:

McGenerous meets his 3 Friends in Vienna. It is Winter and they decide to make a tour through the so called "Bermuda Triangle" (down town). McGenerous wants to invite his friends and Says "be my guests to night,
I'll pay for all of us!"

They start at pub No.1. At the wardrobe they have to leave their coats. Then half of McGenerous' money is spent on beer. When leaving, all the guys take their coats and McGenerous pays 1.- for each. Then they walk over and enter pub No.2. The same procedure: first leaving the coats, then spending half of his money on delicious things.

Vienna is famous for charging only 1.- per coat if you want it back. After leaving Pub No.4 McGenerous' wallet is empty.

Obvious question: what amount of money is needed to start such a tour?
Bonus question: what's the formula for n friends, m pubs and x.- for the
coats?

8)

2003-07-03, 17:38	#1
Wacky Jun 2003 The Texas Hill Country 441₁₆ Posts	A biscuit for the dog Four tramps bought, borrowed, found, or otherwise came into possession of a tin of biscuits. They agreed that they would evenly divide them in the morning. During the night, while the others were asleep, one of them opened the tin and found that if he divided the lot into four equal shares, there would be one left over. To eliminate the uneven division, he fed one to the dog and removed exactly 1/4 of the remainder. Later, a second man did the same. The dog again received a bribe of one biscuit to remain quiet and the man removed exactly 1/4 of the remaining ones. Subsequently, the third and fourth did likewise. In the morning, everyone noticed the reduced number of biscuits in the tin, but said nothing since he had removed some of them. So they divided them equally and found one left over, which again went to the dog. In total, the dog got 5. What is the minimum number of biscuits that were originally in the tin?

2003-07-03, 18:55	#2
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 641 Posts	85 .. yes?

2003-07-03, 19:12	#3
Jwb52z Sep 2002 31E₁₆ Posts	If I understand this, we're not talking about American biscuits, right?

2003-07-03, 19:16	#4
Maybeso Aug 2002 Portland, OR USA 2·137 Posts	There were 1021 biscuits in the tin. That's an impressive sized tin.

2003-07-03, 20:05	#6
Maybeso Aug 2002 Portland, OR USA 422₈ Posts	For N tramps, the ___ (N+1) tin should contain __N___ - (N - 1) biscuits. [edit] For N tramps, the tin should contain N^(N+1) - (N - 1) biscuits. [/edit] An interesting observation, if the 4 tramps would rather not give the dog any biscuits, there would need to be 1024 biscuits in the tin -- 3 more, not less.

2003-07-03, 20:36	#7
NickGlover Aug 2002 Richland, WA 2²·3·11 Posts	Well, I can't read Maybeso's answer (unsupported characters?), but my answer is 4^(N+1) - 3.

2003-07-03, 21:27	#8
Maybeso Aug 2002 Portland, OR USA 2·137 Posts	Sorry about that, I tried to get fancy. For N tramps, the tin should contain N^(N+1) - (N - 1) biscuits. Nick, you have the 4 and the 3 as a constant, but I think they also have to vary with N. (It is nice to see we're thinking along the same lines.) ;)

2003-07-03, 21:54	#9
NickGlover Aug 2002 Richland, WA 10000100₂ Posts	Oh, you are right that my answer is wrong in keeping the 4 and 3. Intuitively, your answer seems right, but I don't have time to verify it right now. I will look at it later this evening if Wackerbarth hasn't confirmed it by then.

2003-07-03, 22:23	#10
Wacky Jun 2003 The Texas Hill Country 441₁₆ Posts	For N tramps, the tin should contain N^(N+1) - (N - 1) biscuits. Spot On, Gentlemen! I'll have to avoid such easy ones ....

2003-07-04, 15:20	#11
koal Nov 2002 Vienna, Austria 51₈ Posts	Had a similar idea years ago: McGenerous meets his 3 Friends in Vienna. It is Winter and they decide to make a tour through the so called "Bermuda Triangle" (down town). McGenerous wants to invite his friends and Says "be my guests to night, I'll pay for all of us!" They start at pub No.1. At the wardrobe they have to leave their coats. Then half of McGenerous' money is spent on beer. When leaving, all the guys take their coats and McGenerous pays 1.- for each. Then they walk over and enter pub No.2. The same procedure: first leaving the coats, then spending half of his money on delicious things. Vienna is famous for charging only 1.- per coat if you want it back. After leaving Pub No.4 McGenerous' wallet is empty. Obvious question: what amount of money is needed to start such a tour? Bonus question: what's the formula for n friends, m pubs and x.- for the coats? 8)