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# neat math riddle

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posted on Mar, 6 2014 @ 12:02 PM
reply to post by saneguy

Why are you so hung up on how much they each paid in the end? Change the riddle from 3 guys to one.

One guy walked into a motel and paid \$30. The hotel clerk said that was too much and gave \$5 back to the bellboy to return to him. Bellboy decided to stiff him \$2 and only handed him back \$3.

Is there a missing dollar when only one person walked in the door? He ended up paying \$27 for the room, as well. In the end, one person or 1,000 people paid \$27, but that has no relevance at all and is what tricks people when they read the riddle. The riddle asks where is the missing dollar, NOT what did each of the guests pay for the room.

Where did each of the 30 bills go?

posted on Mar, 6 2014 @ 12:14 PM
reply to post by ChaoticOrder

This is really above my paygrade... with this indication, can you imagine how much I get paid?

EDIT: This is not a puzzle or riddle, just plain reality.

posted on Mar, 6 2014 @ 01:21 PM
reply to post by CoherentlyConfused

This has been posted here before.

It not even a real maths problem as what is said is just not true

So in the end they each paid \$9 dollars for the room

is not true so its a false premise.

edit on 6-3-2014 by PhoenixOD because: (no reason given)

posted on Mar, 6 2014 @ 04:10 PM

PhoenixOD
reply to post by CoherentlyConfused

This has been posted here before.

It not even a real maths problem as what is said is just not true

So in the end they each paid \$9 dollars for the room

is not true so its a false premise.

edit on 6-3-2014 by PhoenixOD because: (no reason given)

Exactly!

Finally someone get's it!

Korg.

posted on Mar, 6 2014 @ 05:15 PM
reply to post by Korg Trinity

Please explain. Without any math, since it's not really a math problem. I mean, I read your replies before, which contained all kinds of irrelevant math problems, so .... I really am confused now.
What did I miss here? Something has gone over my head. Thanks.

posted on Mar, 6 2014 @ 05:37 PM

PhoenixOD
reply to post by CoherentlyConfused

This has been posted here before.

It not even a real maths problem as what is said is just not true

So in the end they each paid \$9 dollars for the room

is not true so its a false premise.

edit on 6-3-2014 by PhoenixOD because: (no reason given)

Actually, it's true. In the end each paid \$9. The problem is that each was only charged \$8-1/3 by the hotel. That means the bellboy took \$2/3 from each guest. That's where he got his \$2 from, 3 x 2/3 = 2.

posted on Mar, 6 2014 @ 06:16 PM
reply to post by DenyObfuscation

Actually, it's true. In the end each paid \$9. The problem is that each was only charged \$8-1/3 by the hotel. That means the bellboy took \$2/3 from each guest. That's where he got his \$2 from, 3 x 2/3 = 2.

Nope..they each paid 9.3333 recurring.

\$30 - \$5 = \$25
\$25 /3 =\$8.33333 recuring
So each paid \$8.33333 recurring

belboy returned \$3
\$3 / 3 = \$1
\$8.33333+\$1 = \$9.33333 recurring
so each paid \$9.33333 recurring

This can be proved by adding it all back together:
\$9.33333 x 3 = \$28
\$28 + \$2 that the bellboy took is \$30

edit on 6-3-2014 by PhoenixOD because: (no reason given)

posted on Mar, 6 2014 @ 06:35 PM
reply to post by PhoenixOD

Nope..they each paid 9.3333 recurring.

That can't be correct. Originally each paid 10. Each received 1 back reducing the amount paid to 9 each. They ended up paying 27 even though were only charged 25. The bellboy got his 2 by keeping \$2/3 from each guest's "change" returned.

Originally \$30 charged and paid. Charge was reduced to \$25. That's \$8-1/3 each. Each was due a refund of \$1-2/3 but only received \$1 reducing their payment to \$9 each.

30-25=5 5-3=2

posted on Mar, 6 2014 @ 06:46 PM
reply to post by DenyObfuscation

Except the way you are trying to work it out there is \$1 missing which is impossible

The answer is they paid \$9.3333333 there simply is no other correct answer.

Why not go a to a dedicated maths forum and try to tell them your method and report back what they say.

posted on Mar, 6 2014 @ 06:48 PM
reply to post by PhoenixOD

So they originally paid \$10.333333333333 each?

10.333333333333 - 1 = 9.3333333333333

posted on Mar, 6 2014 @ 06:51 PM
reply to post by DenyObfuscation

here is what an expert has to say

Dr. Rothman numbers the dollars:

Let's give each of the \$30 a number from 1-30, keep track of each individual dollar, and see how the problem works.

Dollars numbered 1-30 are given to the manager. Then he wants to give \$5 back, so he keeps the dollars numbered 1-25, and gives numbers 26-30 to the bellboy in the form of a five dollar bill. The bellboy splits up the five to get 5 one's: numbers 26, 27, 28, 29 and 30. He gives numbers 26, 27 and 28 to the customers and keeps numbers 29 and 30 for himself.

there was \$30 and the bellboy took \$2 which = 28

\$28/3 which is \$9.3333333 each.

So now i have presented 2 way which account for all the money. Quite simply if your method can not account for ALL the dollars then it is incorrect because \$1 cant just vanish

edit on 6-3-2014 by PhoenixOD because: (no reason given)

posted on Mar, 6 2014 @ 06:55 PM
reply to post by PhoenixOD

here is what an expert has to say

That's great, the expert says exactly what I've said. Where does he mention them paying 9.3333 each? He doesn't because they didn't.

9.333333 + 1 = 10.3333333. That didn't happen.

ETA:

there was \$30 and the bellboy took \$2 which = 28/3 which is \$9.3333333 each.

That's Mafia accounting.

30 - 25 = 5. 5 - 3 = 2
edit on 6-3-2014 by DenyObfuscation because: (no reason given)

posted on Mar, 6 2014 @ 06:57 PM
reply to post by DenyObfuscation

The expert says that dollars 1-25 and dollars 26,27,28 go to the people and 29, 30 go to the bellboy.

yet again 28/3 =9.3333333

If you cant account for all the dollars your method is flawed

edit on 6-3-2014 by PhoenixOD because: (no reason given)

posted on Mar, 6 2014 @ 07:02 PM

PhoenixOD
reply to post by DenyObfuscation

The expert says that dollars 1-25 and dollars 26,27,28 go to the people and 29, 30 go to the bellboy.

yet again 28/3 =9.3333333

If you cant account for all the dollars your method is flawed

edit on 6-3-2014 by PhoenixOD because: (no reason given)

That's the problem. 1-25 go to the hotel. 26-28 go back to the guests. 29 & 30 go to the crooked bellboy.

posted on Mar, 6 2014 @ 07:11 PM
reply to post by DenyObfuscation

Hmmm... i think i understand this better now :

They actually paid 8.333333 to the hotel as 25 is what remained in the till.

But because the bellboy ripped them off they parted with \$9.

The trick is to realize that the \$2 has to be subtracted from the \$27, not added to it.

good call and all money accounted for

edit on 6-3-2014 by PhoenixOD because: (no reason given)

posted on Mar, 6 2014 @ 07:18 PM
reply to post by PhoenixOD

You got it. Now the trick is to see if anyone can get Korg to come around on this!

posted on Mar, 6 2014 @ 07:21 PM
reply to post by DenyObfuscation

I knew there was a false premise in there somewhere, with maths riddles like this there always is. This time the false premise was that you have to add the \$2 to the \$27. Had me fooled for a while lol

edit on 6-3-2014 by PhoenixOD because: (no reason given)

posted on Mar, 6 2014 @ 07:26 PM
What they paid in the end is irrelevant. How many people there were is irrelevant. How they split it up is irrelevant.

original amount paid: \$30.

Hotel kept \$25, bellboy stole \$2 and they got a \$3.00 refund.

25+2+3=30.

People have been saying that for 3 pages now. Someone even said that back on page 1.

posted on Mar, 6 2014 @ 07:31 PM

PhoenixOD
reply to post by DenyObfuscation

The expert says that dollars 1-25 and dollars 26,27,28 go to the people and 29, 30 go to the bellboy.

yet again 28/3 =9.3333333

If you cant account for all the dollars your method is flawed

edit on 6-3-2014 by PhoenixOD because: (no reason given)

You are missing the point, there is \$3 left, say they each buy a slice of pizza with that \$3, the bell boy gets \$2 and the hotel gets \$25 = \$30 dollars.

posted on Mar, 6 2014 @ 07:37 PM
reply to post by seaez

Yep i agree, i noticed that after posting and corrected myself in a later post

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