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Author Topic: Math oddities  (Read 8900 times)

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Offline KurzykTopic starter

Math oddities
« on: April 06, 2011, 06:16:45 PM »
Someone pointed something interesting out to me today.

If you take how old you'll be on your birthday this year and add it to the last two numbers of your birthyear, it will equal 111.

So for me, I was born 5/4/1972.

Ill be 39 this year + 72 = 111.

Try it. Everyone so far equals 111 and it seems to just be for this year.
« Last Edit: April 09, 2011, 11:03:28 AM by Kurzyk »

Offline Jude

Re: Math oddity
« Reply #1 on: April 06, 2011, 06:30:10 PM »
"If you take how old you'll be on your birthday this year and add it to the last two numbers of your birthyear, it will equal 111."

If you are 10 or younger this actually doesn't hold up.  Only for people born in the last century does it pan out.  However, in all circumstances you will get 11 or 111, and that's just because you're adding when you were born + how long you've been alive to get the current year, 2011, basically.  The extra 100 comes in because we're truncating the first two digits (19) from your birth year, so that 100 doesn't add to it to make it (20).
« Last Edit: April 06, 2011, 06:31:11 PM by Jude »

Offline KurzykTopic starter

Re: Math oddity
« Reply #2 on: April 06, 2011, 07:35:33 PM »
Cool thank you. I was wondering how that worked.  :-)

Offline Wolfy

Re: Math oddity
« Reply #3 on: April 09, 2011, 03:17:36 AM »
I was born in 90.

This year I turned 21, which does add up to 111.

However, next year I'll be 22, and that adds up to 112.

:/

Offline Oniya

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Re: Math oddity
« Reply #4 on: April 09, 2011, 10:26:59 AM »
But next year will be 2012.

[Y]ou're adding when you were born + how long you've been alive to get the current year, 2011, basically.  The extra 100 comes in because we're truncating the first two digits (19) from your birth year, so that 100 doesn't add to it to make it (20).

Offline KurzykTopic starter

Re: Math oddities
« Reply #5 on: April 09, 2011, 11:03:09 AM »
Ok here's another one. I found it on a math oddities website:

Three guys walk into an hotel and ask for a room with three beds. They just want to stay for one night. The man at the hotel tells them that the cost will be 30 dollars. They pay 10 dollars each and walk upstairs to their room. After some minutes the man knocks at their door, apologizing for a mistake. He made them pay the high season price and not the low season one, and since now it's the low season, he must give them 5 dollars back.

They are happy and appreciate the honesty of the man, so they tell him: "from these 5 dollars we will just take back 1 dollar each and we will give you the remaining 2 dollars as a tip for your honesty". The man thanks the guys and walks out.

Ok, they paid 10 dollars each (30 in total). They take back 1 dollar each (3 dollars in total). In this way they have paid 10-1=9 dollar each. So in total they have now paid 9*3=27 dollars. Plus the 2 dollars tip given to the man we get 27+2=29 dollars. Where is the missing dollar?


In reading this, my impression is that the dollar isn't really missing. Is it? I mean they gave 30, it was supposed to be 25, so they received 5 back. They kept 3 and gave 2.

Yet looking at that math it does look like 1 is missing.
« Last Edit: April 09, 2011, 11:05:56 AM by Kurzyk »

Offline Oniya

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Re: Math oddities
« Reply #6 on: April 09, 2011, 11:41:02 AM »
Actually, they paid 27 to the hotel, 25 of which went to the room, and 2 of which went to the man.  The 2 is subtracted from the 27 to get the price of the room.

Offline KurzykTopic starter

Re: Math oddities
« Reply #7 on: April 09, 2011, 11:44:00 AM »
Oh.. so they initially gave him 30, got 3 back and gave him 2, leaving 1 missing.

But if the price of the room is 25 off season, and he gave them 5 back, does that mean that the innkeeper kept the extra dollar?

Offline Oniya

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Re: Math oddities
« Reply #8 on: April 09, 2011, 11:46:55 AM »
No.  The way it goes is $30 (what they paid initially) - $3 (what they got back) = $25 (what the room cost) + $2 (what the manager got).

Offline KurzykTopic starter

Re: Math oddities
« Reply #9 on: April 09, 2011, 11:50:14 AM »
Right. That's what I was initially thinking. So then there is no extra dollar? They paid 30, got back 3 and gave him 2. It still totals 30.

Offline Oniya

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Re: Math oddities
« Reply #10 on: April 09, 2011, 11:54:30 AM »
No extra dollar.  If you think of the $5 as five $1 bills, two of them never leave the possession of the hotel manager.

Offline NotoriusBEN

Re: Math oddities
« Reply #11 on: April 12, 2011, 10:18:49 PM »
yep, what oniya said.

plus, that money swap with the hotel manager is one of the "oldest" scams in the book.  When I worked in retail years ago, I fell for it a couple times.

If you work in retail, do like they do in Vegas. Keep all money on the table and only transition it when the final amounts are known. Beware the person talking a lot of different numbers and asking or trying to change money mid transition. :P

Offline DreamlandDenizen

Re: Math oddities
« Reply #12 on: April 12, 2011, 10:38:41 PM »
"Ok, they paid 10 dollars each (30 in total). They take back 1 dollar each (3 dollars in total). In this way they have paid 10-1=9 dollar each. So in total they have now paid 9*3=27 dollars. Plus the 2 dollars tip given to the man we get 27+2=29 dollars. Where is the missing dollar?"

The problem with the math is you should not add the two dollars to the 27 near the end. You say they take 3 dollars back (not five) so the 2 dollar difference stays with the manager the whole time and is thus already accounted for. So leaving out that error you're left with 27, the amount they actually paid in the end, and to get back to the 30 you just add the $3 they took back.
« Last Edit: April 12, 2011, 10:42:15 PM by DreamlandDenizen »

Offline ReflectManSmile

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Re: Math oddities
« Reply #13 on: April 13, 2011, 03:32:53 PM »
This last oddity reminds me the monty hall problem, in where you have a result that seems absurd, but which is demonstrable.

Offline KurzykTopic starter

Re: Math oddities
« Reply #14 on: April 13, 2011, 05:37:35 PM »
Interesting example Reflect. Pulling it out from the article:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Wouldn't the chances be equal between 1 and 2? So knowing 3 wouldn't be an advantage regarding your choices between 1 and 2.

Offline Remiel

Re: Math oddities
« Reply #15 on: April 13, 2011, 06:22:11 PM »
Ok here's another one. I found it on a math oddities website:

Three guys walk into an hotel and ask for a room with three beds. They just want to stay for one night. The man at the hotel tells them that the cost will be 30 dollars. They pay 10 dollars each and walk upstairs to their room. After some minutes the man knocks at their door, apologizing for a mistake. He made them pay the high season price and not the low season one, and since now it's the low season, he must give them 5 dollars back.

They are happy and appreciate the honesty of the man, so they tell him: "from these 5 dollars we will just take back 1 dollar each and we will give you the remaining 2 dollars as a tip for your honesty". The man thanks the guys and walks out.

Ok, they paid 10 dollars each (30 in total). They take back 1 dollar each (3 dollars in total). In this way they have paid 10-1=9 dollar each. So in total they have now paid 9*3=27 dollars. Plus the 2 dollars tip given to the man we get 27+2=29 dollars. Where is the missing dollar?


In reading this, my impression is that the dollar isn't really missing. Is it? I mean they gave 30, it was supposed to be 25, so they received 5 back. They kept 3 and gave 2.

Yet looking at that math it does look like 1 is missing.

Whew!  This one threw me for a while.  I knew, obviously, that the dollar really wasn't "missing", but I couldn't for the life of me explain why.  I think I've figured it out, however.  Here goes:

The amount that the hotel offers to refund is a red herring--that is, it's a completely misleading number.  As Oniya referred to, the amount that actually matters is the total amount actually paid to the hotel (including the tip).

To illustrate, let's say our three fellows book rooms at the same hotel the next season.  Once again, they pay the hotel $10 each, for a grand total of $30.  This time, however, let's say that our hotel manager is so grateful for their business (and perhaps a little drunk) that he offers to refund $20.  Once again, our generous travelers refuse, accepting only $1 back each and giving him the other $17 as a tip.   So now our travelers have paid a grand total of  $44 ($10 - 1 = 9 dollars each x 3 = $27 + the $17 tip = $44).  Right?

Obviously that is not the case.  In both examples, the travelers have paid $27 (10 initial - 1 back = $9 each x 3 = $27).  The only difference is that, in the first example, the manager made a $2 (or $5 - 3) tip, while in the second he made $17 (or $20 - 3).   The amount of the refund only factors into how much tip the manager gets, and has nothing to do with what the guys pay.

Offline Oniya

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Re: Math oddities
« Reply #16 on: April 13, 2011, 06:24:21 PM »
Interesting example Reflect. Pulling it out from the article:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Wouldn't the chances be equal between 1 and 2? So knowing 3 wouldn't be an advantage regarding your choices between 1 and 2.

Except that the chances weren't equal when you initially picked.  Also pulled from the article:

Code: [Select]
Door 1 Door 2 Door 3 result if switching result if staying
Car    Goat   Goat    Goat                    Car
Goat    Car    Goat    Car                     Goat
Goat    Goat   Car     Car                     Goat

The results columns are assuming that you picked Door 1, and Monty revealed one of the goats.

Offline KurzykTopic starter

Re: Math oddities
« Reply #17 on: April 13, 2011, 06:27:57 PM »
Its true that the chances were different when the initial choice of door 1 was chosen, but the question was whether it was an advantage in changing my choice. Its still 50/50 as to whether its 1 or 2.

Offline Remiel

Re: Math oddities
« Reply #18 on: April 13, 2011, 06:28:17 PM »
Also, the movie 21 has a pretty good explanation of the problem and the solution.

"21" explains the Monty Hall problem

Offline Jude

Re: Math oddities
« Reply #19 on: April 14, 2011, 03:39:33 PM »
Its true that the chances were different when the initial choice of door 1 was chosen, but the question was whether it was an advantage in changing my choice. Its still 50/50 as to whether its 1 or 2.
The problem with that logic is that Monty Hall knows which door the car is behind and cannot pick it to be the odd one out by the rules of the game.  Look at it this way:

Lets say you choose door 1 then break it down into cases from there.  There are 3 possibly situations, and each is defined by the door the car is behind.

Door 1 Car:  his choice of which door to exclude is arbitrary so no additional information is given by it.
Door 2 Car:  he actually has no choice on which door to exclude, it has to be number 3 because 2 has the car.  Thus he is exposing door 2 as the car.
Door 3 Car:  he actually has no choice on which door to exclude, it has to be number 2 because 3 has the car.  Thus he is exposing door 3 as the car.

In two out of three situations his actions give away what door the car is behind, so changing your choice based on his actions will more often than not lead to a better outcome.
« Last Edit: April 14, 2011, 03:41:35 PM by Jude »

Offline samantha_cs

Re: Math oddities
« Reply #20 on: May 19, 2011, 01:24:10 PM »
I find it easier to think about the probability shift if you take it to a ridiculous extreme.

Let's say there are 100 doors, with 1 prize behind one of them. You pick one door. Monty Hall opens 98 doors you didn't pick and shows there is nothing behind any of them.

You now have the option of changing doors.

Isn't it pretty obvious that you should switch doors in this example?

Offline Shjade

Re: Math oddities
« Reply #21 on: May 19, 2011, 02:17:20 PM »
I can't really see that the 100 to 1 example is more/less obvious than the 3 to 1 case. In either circumstance the rules end up the same: either you picked the door with the car behind it and the opened doors are arbitrary or you didn't pick the car and the door remaining has the car behind it. Works the same whether there are 3 doors or 3000. Right?

Offline WhiteyChan

Re: Math oddities
« Reply #22 on: May 19, 2011, 04:34:18 PM »
Changing slightly (read: a lot), another mathematical oddity for you, a lot more abstract but nonetheless amusing if you can understand it.

It is completely possibly for 1 + 1 = 0 to be true.

...

If you are in the Set F2, that is.

Offline samantha_cs

Re: Math oddities
« Reply #23 on: May 19, 2011, 05:24:46 PM »
I can't really see that the 100 to 1 example is more/less obvious than the 3 to 1 case. In either circumstance the rules end up the same: either you picked the door with the car behind it and the opened doors are arbitrary or you didn't pick the car and the door remaining has the car behind it. Works the same whether there are 3 doors or 3000. Right?
Yes, but the odds you picked the right door on your initial try are much, much smaller. Ignoring the math, which doesn't care if it's 3 or 3000, your 'common sense' should tell you to switch doors in the 100 (or 1000 or 10,000) to 1 case.

Offline Shjade

Re: Math oddities
« Reply #24 on: May 19, 2011, 05:29:26 PM »
Did you just say "ignoring the math" in a thread about math? ;p

Still, fair point.