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## Question

"The three of us made some bets," someone said.

"First, A won from B as much as A had originally. Next, B won from C as much as B then had left. Finally, C won from A as much as C then had left. We ended up having equal amounts of money. I began with 50 cents."

Who is the speaker? A, B, or C?

Note: A basic understanding of algebra is required for a solution.

Good Luck!

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Oh, this one's nice.

The speaker is B. They started with 30, 50, and 40 cents.

The speaker cannot be A, because A's amount of money must be a multiple of 3. The speaker cannot be C, because then A would have to start with a negative amount of money.

Explanation:

We'll solve it with equations. Call the starting sums A, B, and C, and the final amount X. We'll measure in cents.

Either A, B, or C = 50 (given)

What A had after the first bet: 2A

What B had after the first bet: B - A

What B had after the second bet: 2(B - A)

What C had after the second bet: C - (B - A) = A - B + C

What C had after the third bet: 2(A - B + C)

What A had after the third bet: 2A - (A - B + C) = A + B - C

So we have: 2A - 2B + 2C = A + B - C = 2B - 2A

Let's break it into two equalities:

2A - 2B + 2C = A + B - C

becomes

A - 3B + 3C = 0 => A = 3(B - C)

It's impossible for A to be 50 cents! It's not a multiple of 3!

And

A + B - C = 2B - 2A

becomes

C = 3A - B

And

2A - 2B + 2C = 2B - 2A

becomes

C = 2(B - A)

Time for brute force. I'm sure a more efficient solution exists. But let's suppose B is 50 cents. Then:

3A - 50 = C

and

100 - 2A = C

Multiplying the bottom equation by 3 and to top by 2, we get:

6A - 100 = 2C

300 - 6A = 3C

200 = 5C

C = 40

What does this make A?

100 - 2A = 40

A = 30.

Check to see if this works:

First bet:

A: 30 -> 60

B: 50 -> 20

Second bet:

B: 20 -> 40

C: 40 -> 20

Third bet:

C: 20 -> 40

A: 60 -> 40

It checks!

Finally, to make sure there isn't a second solution, let's suppose C had 50 cents to start with.

C = 2(B - A)

A = 3(B - C)

50 = 2B - 2A -> 2A = 2B - 50

A = 3B - 150

Double the bottom equation:

2A = 6B - 150

2A = 2B - 50

Subtract:

0 = 4B - 100

B = 25

Now we find A:

A = 3B - 150

Aha. A cannot be a negative value. Nobody can start with negative money in this particular situation, as we couldn't talk about "what A had left" if they allowed playing in debt.

Edited by WitchOfDoubt
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Oh, this one's nice.

The speaker is B. They started with 30, 50, and 40 cents.

The speaker cannot be A, because A's amount of money must be a multiple of 3. The speaker cannot be C, because then A would have to start with a negative amount of money.

Explanation:

We'll solve it with equations. Call the starting sums A, B, and C, and the final amount X. We'll measure in cents.

Either A, B, or C = 50 (given)

What A had after the first bet: 2A

What B had after the first bet: B - A

What B had after the second bet: 2(B - A)

What C had after the second bet: C - (B - A) = A - B + C

What C had after the third bet: 2(A - B + C)

What A had after the third bet: 2A - (A - B + C) = A + B - C

So we have: 2A - 2B + 2C = A + B - C = 2B - 2A

Let's break it into two equalities:

2A - 2B + 2C = A + B - C

becomes

A - 3B + 3C = 0 => A = 3(B - C)

It's impossible for A to be 50 cents! It's not a multiple of 3!

And

A + B - C = 2B - 2A

becomes

C = 3A - B

And

2A - 2B + 2C = 2B - 2A

becomes

C = 2(B - A)

Time for brute force. I'm sure a more efficient solution exists. But let's suppose B is 50 cents. Then:

3A - 50 = C

and

100 - 2A = C

Multiplying the bottom equation by 3 and to top by 2, we get:

6A - 100 = 2C

300 - 6A = 3C

200 = 5C

C = 40

What does this make A?

100 - 2A = 40

A = 30.

Check to see if this works:

First bet:

A: 30 -> 60

B: 50 -> 20

Second bet:

B: 20 -> 40

C: 40 -> 20

Third bet:

C: 20 -> 40

A: 60 -> 40

It checks!

Finally, to make sure there isn't a second solution, let's suppose C had 50 cents to start with.

C = 2(B - A)

A = 3(B - C)

50 = 2B - 2A -> 2A = 2B - 50

A = 3B - 150

Double the bottom equation:

2A = 6B - 150

2A = 2B - 50

Subtract:

0 = 4B - 100

B = 25

Now we find A:

A = 3B - 150

Aha. A cannot be a negative value. Nobody can start with negative money in this particular situation, as we couldn't talk about "what A had left" if they allowed playing in debt.

i did the exact same thing to obtain the equation but i solved the equations to get a/3=b/5=c/4 and concluded b=50 cents hence the speaker
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Hey great job guys, but dheeraj54, could you explain that technique of yours?

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let the initial amounts they have be respectively a,b & c

A~~~~~~~~~~~~~~~B~~~~~~~~~~~~~~~~~C

a ~~~~~~~~~~~~~~ b~~~~~~~~~~~~~~~~~ c

(a+a) ~~~~~~~~~~~ (b-a)~~~~~~~~~~~~~~~ c

2a ~~~~~~~~~~ (b-a)+(b-a)~~~~~~~~~~ c-(b-a)

2a-(a+c-b) ~~~~~~~~ 2(b-a)~~~~~~~~~ (a+c-b)+(a+c-b)

a+b-c ~~~~~~~~~~~ 2b-2a ~~~~~~~~~~~~ 2a+2c-2b

Now, as the final amounts are equal,, after equating the three.we get : a= 3b/5 , c=4b/5

putting b=50 ,we get a=30, b=50, c=40

B is the speaker

Edited by Arjit Saraswat

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