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There were a group of people sitting in a row. John, the rightmost guy, put certain amount of money on the table and the guy who sat to the left of him put one dollar less on the table. This pattern continued upto the leftmost guy. Then John gave one dollar to the man next to him, and that man gave two dollars to the guy on his left and so on. This process went on till the leftmost guy. At this time he has four times the money to his right person.

Now my questions are how many men were there along with John and how much money had the leftmost guy have?

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At this time he has four times the money to his right person.

brhan, can you please clarify this for me? Does it mean the leftmost guy has 4x the money of the person directly to his right?

Edited by itachi-san

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brhan, can you please clarify this for me? Does it mean the leftmost guy has 4x the money of the person directly to his right?

exactly ... the money was being collected by the leftmost guy. And hence the leftmost guy has four times the money of the guy to his right. Sorry, I just saw it late.

Edited by brhan

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1 2 3 4 ...(or) 1 2 4 8 16....

The first one. Everyone passes the collected money to his left adding one dollar. So, it goes 1 2 3 4 ...

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Ok, I'm seeing two totally independent events.

1. they each put money on the table, John putting x dollars and person to his left putting x-1 dollars until it reaches end (so there is apparently a pile of money on the table).

2. they each pass money down, John passing $1 and person to right pass $2 and so on all the way until end (so person on left has the, um, donation cup).

How does first part relate to second?

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Ok, I'm seeing two totally independent events.

1. they each put money on the table, John putting x dollars and person to his left putting x-1 dollars until it reaches end (so there is apparently a pile of money on the table).

2. they each pass money down, John passing $1 and person to right pass $2 and so on all the way until end (so person on left has the, um, donation cup).

How does first part relate to second?

The first part (1) is the money they had. They didn't pass the money. To put it in another way, initially each person had money one dollar less from rightmost to leftmost. If there were three guys, say A, B and C from right to left and A had $3, then B would have $2 and C would have $1.

In the second "step" (2), A gave $1 to B, B gave $2 to C. As C was the leftmost guy, there was no one to give to. So he mixed it with the money initially he had and that money was four times the amount of money B had.

Hope that makes sense to you.

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There are four people at the table.

Initially:

a = 4, b= 3, c= 2, d = 1

After passing to the left:

a=3, b= 2, c= 1, d= 4

Edited by Nikyma

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There are four people at the table.

Initially:

a = 4, b= 3, c= 2, d = 1

After passing to the left:

a=3, b= 2, c= 1, d= 4

You're not accounting for the first part of the puzzle when they each put money down on the table.

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There were a group of people sitting in a row. John, the rightmost guy, put certain amount of money on the table and the guy who sat to the left of him put one dollar less on the table. This pattern continued upto the leftmost guy. Then John gave one dollar to the man next to him, and that man gave two dollars to the guy on his left and so on. This process went on till the leftmost guy. At this time he has four times the money to his right person.

Now my questions are how many men were there along with John and how much money had the leftmost guy have?

5 men

A(john gives) $1 B gives $2 C gives $3 D gives $4 E collect $4 (from D)

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Let , x= John's initial amount of money , in $ .

n=total number of people(including John)

Then the seating arrangement is like:

Order of seating(L to R) : n , n-1 , ...... , 3 , 2 , 1(John)

John, the rightmost guy, put certain amount of money on the table and the guy who sat to the left of him put one dollar less on the table. This pattern continued upto the leftmost guy.

Initial amount of money :

1(John) = x

2 = x-1

3 = x-2

.

.

.

.

n-1 = x-(n-2)

n = x-(n-1) ----> Leftmost person

Then John gave one dollar to the man next to him, and that man gave two dollars to the guy on his left and so on. This process went on till the leftmost guy.

Final amount of money:

1(John) = x-1

2 = [x-1+1] - 2 = x-2

3 = [x-2+2] - 3 = x-3

.

.

.

.

n-1 = [x-(n-2)+(n-2)] - (n-1) = x-(n-1)

n = [x-(n-1)] + (n-1) = x

At this time the leftmost person has four times the money to his right person.

x = 4[x-(n-1)]

Solving , we get , an equation ,

3x = 4n - 4

Substituting , n=4 , R.H.S = 12

Substituting , x=4 , L.H.S = 12

Therefore , L.H.S = R.H.S

Now my questions are how many men were there along with John and how much money had the leftmost guy have?

3 men were along with John and the leftmost guy had $4. Maybe your topic title was a clue.;)

Hope it is correct. :)

Edited by grey cells

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Topic title is a clue

double money - $2

maths lovers = 4 Possible a on math lover there too.

Or no one is right yet!

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Hope it is correct. :)

Let , x= John's initial amount of money , in $ .

n=total number of people(including John)

Then the seating arrangement is like:

Order of seating(L to R) : n , n-1 , ...... , 3 , 2 , 1(John)

John, the rightmost guy, put certain amount of money on the table and the guy who sat to the left of him put one dollar less on the table. This pattern continued upto the leftmost guy.

Initial amount of money :

1(John) = x

2 = x-1

3 = x-2

.

.

.

.

n-1 = x-(n-2)

n = x-(n-1) ----> Leftmost person

Then John gave one dollar to the man next to him, and that man gave two dollars to the guy on his left and so on. This process went on till the leftmost guy.

Final amount of money:

1(John) = x-1

2 = [x-1+1] - 2 = x-2

3 = [x-2+2] - 3 = x-3

.

.

.

.

n-1 = [x-(n-2)+(n-2)] - (n-1) = x-(n-1)

n = [x-(n-1)] + (n-1) = x

At this time the leftmost person has four times the money to his right person.

x = 4[x-(n-1)]

Solving , we get , an equation ,

3x = 4n - 4

Substituting , n=4 , R.H.S = 12

Substituting , x=4 , L.H.S = 12

Therefore , L.H.S = R.H.S

Now my questions are how many men were there along with John and how much money had the leftmost guy have?

3 men were along with John and the leftmost guy had $4. Maybe your topic title was a clue.;)

Nikyama and grey cells have got it. Great.

Just to include my comment, grey cells has got 3x=4n-4, where x is John's original money and 'n' their number including John.

We solve for 'n' : n=3x/4 + 1 ... and 'n' should be an integer. When x=4 we have n=4. Other solutions ...

x=8 and n=7

x=12 and n=10 ... etc.

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There were 8 people and the leftmost guy had 8$ after he collect the money,initialy he had 1$.

There were a group of people sitting in a row. John, the rightmost guy, put certain amount of money on the table and the guy who sat to the left of him put one dollar less on the table. This pattern continued upto the leftmost guy. Then John gave one dollar to the man next to him, and that man gave two dollars to the guy on his left and so on. This process went on till the leftmost guy. At this time he has four times the money to his right person.

Now my questions are how many men were there along with John and how much money had the leftmost guy have?

Edited by ash013

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There were 8 people and the leftmost guy had 8$ after he collect the money,initialy he had 1$.

The money is correct ... but there were 7 people. You may check my previous post for the explanation.

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Little late to the party, but aren't there multiple answers, whenever the ratio of John to Last person is 4:1.

Example- John has 4 dollars and there are 4 people, Last guy has a dollar. Also, John has 8 dollars and there are 7 people, Last guy has 2 dollars. And so on...

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