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DieCommie

Registered: 12/11/03
Posts: 29,258
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Quote:
DividedQuantum said: PROBLEM #15
Quote:
In this multiplication problem each P stands for a prime digit (2,3,5, or 7). Solve the problem. The answer is unique.
P P P P P ----------- P P P P P P P P -------------- P P P P P
I am always terrible at these types of brain teasers... but this one really doesn't make sense to me. Are the least significant digits supposed to line up? Is this two multiplications happening?
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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Re: mathematical brain teaser thread [Re: DieCommie]
#21941295 - 07/14/15 09:39 AM (8 years, 6 months ago) |
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Quote:
DieCommie said:
Quote:
DividedQuantum said: Look, nevermind.
This is beginning to be a much bigger deal than it ever should have been.
secondorder, please, by all means, solve the problems however you want to. In the future, it would be better if you provided your thought process with the final answer -- whatever it is. That will be more than acceptable, I assure you. Elegant mathematics are not necessary. They're nice, but they were never necessary. Showing how you got there is important, though.
No more!
I'm still interested in the solution though...
Elegant or not, I thought that there might be some way to algorithmically solve the system with restraints of being an integer and being less than 100. I tried to solve it myself and quickly realized that there was not enough equations to solve for the variables. Does setting up the equation like the answer showed help guide in the guess and check?
edit - Let me dig in a little because I think you can take it... Quote:
Just putting the answer up there with no math or logic, correct or not, is lame.
Copying the answer from somewhere else without understanding enough to know that it lacks logic and math is lame. 
Oh I can take that. Yeah, I'm just going off of some of Martin Gardner's old material for Scientific American. I'll be the first to admit that I am not analyzing these to any significant degree, so I actually appreciate your feedback. It's interesting.
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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Re: mathematical brain teaser thread [Re: DieCommie]
#21941310 - 07/14/15 09:43 AM (8 years, 6 months ago) |
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Quote:
DieCommie said:
Quote:
DividedQuantum said: PROBLEM #15
Quote:
In this multiplication problem each P stands for a prime digit (2,3,5, or 7). Solve the problem. The answer is unique.
P P P P P ----------- P P P P P P P P -------------- P P P P P
I am always terrible at these types of brain teasers... but this one really doesn't make sense to me. Are the least significant digits supposed to line up? Is this two multiplications happening?
Yeah I was worried it might be a little confusing. Actually, it's just an example of longhand multiplication. So, the first four P's under the first line are the result of multiplying the top line by the ones digit (in PP), and the second four P's (moved to the right by an implied zero) are the result of the tens digit. So (hint) actually, the bottom line of five P's is the answer of multiplying the top two lines. In other words, PPP x PP = PPPPP. PPPP + PPPP = PPPPP.
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Sun King



Registered: 02/15/14
Posts: 4,069
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Re: mathematical brain teaser thread [Re: secondorder]
#21941360 - 07/14/15 09:55 AM (8 years, 6 months ago) |
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Quote:
secondorder said: Who are you to tell us what is and what isn't fun?
Damn fun nazis.
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night_shift
Stranger

Registered: 01/02/15
Posts: 226
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Re: mathematical brain teaser thread *DELETED* [Re: DividedQuantum] 1
#21946353 - 07/15/15 08:37 AM (8 years, 6 months ago) |
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Post deleted by night_shift
Reason for deletion: [delete]
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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Re: mathematical brain teaser thread [Re: night_shift]
#21946575 - 07/15/15 09:52 AM (8 years, 6 months ago) |
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Quote:
night_shift said: 7 7 5 3 3 ----------- 2 3 2 5 2 3 2 5 -------------- 2 5 5 7 5
  
Correct! Very well done!
This problem is perhaps best approached by searching first for all three-digit numbers composed of prime digits that yield four prime digits when multiplied by a prime. There are only four:
775 x 3 = 2325 555 x 5 = 2775 755 x 5 = 3775 325 x 7 = 2275
No three-digit number has more than one multiplier; therefore the multiplier in the problem must consist of two identical digits. Thus there are only four possibilities that need to be tested.
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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PROBLEM #16:
Quote:
Ask spectator A to jot down any three-digit number, and then to repeat the digits in the same order to make a six-digit number (e.g. 394,394). Without showing you the number, A is then to pass the sheet of paper to spectator B, who is requested to divide the number by 7.
"Don't worry about the remainder, " you tell B, "because there won't be any." B is surprised to discover that you are right. Without telling you the result, B passes the number to spectator C, who is told to divide it by 11. Once again, you state that there will be no remainder, and this also proves correct.
Direct spectator D to divide the last result by 13. Again, the division comes out even. This final result is written on a slip of paper that is folded and handed to you. Without opening it you pass it on to A.
"Open this," you tell A, "and you will find your original three-digit number."
Prove that the trick cannot fail to work regardless of the original digits chosen.
-------------------- Vi Veri Universum Vivus Vici
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Sun King



Registered: 02/15/14
Posts: 4,069
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Quote:
DividedQuantum said: PROBLEM #16:
Quote:
Ask spectator A to jot down any three-digit number, and then to repeat the digits in the same order to make a six-digit number (e.g. 394,394). Without showing you the number, A is then to pass the sheet of paper to spectator B, who is requested to divide the number by 7.
"Don't worry about the remainder, " you tell B, "because there won't be any." B is surprised to discover that you are right. Without telling you the result, B passes the number to spectator C, who is told to divide it by 11. Once again, you state that there will be no remainder, and this also proves correct.
Direct spectator D to divide the last result by 13. Again, the division comes out even. This final result is written on a slip of paper that is folded and handed to you. Without opening it you pass it on to A.
"Open this," you tell A, "and you will find your original three-digit number."
Prove that the trick cannot fail to work regardless of the original digits chosen.
It's witchcraft. I don't know how to prove witchcraft, other than writing a computer program and I'm too lazy for that.
I'll just trust you that it works.
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OrgoneConclusion
Blue Fish Group



Registered: 04/01/07
Posts: 45,414
Loc: Under the C
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Re: mathematical brain teaser thread [Re: Sun King]
#21948612 - 07/15/15 05:50 PM (8 years, 6 months ago) |
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Quote:
Sun King said:
Quote:
secondorder said: Who are you to tell us what is and what isn't fun?
Damn fun nazis.
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Sun King



Registered: 02/15/14
Posts: 4,069
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I bet that's a good book.
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OrgoneConclusion
Blue Fish Group



Registered: 04/01/07
Posts: 45,414
Loc: Under the C
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Re: mathematical brain teaser thread [Re: Sun King]
#21948696 - 07/15/15 06:08 PM (8 years, 6 months ago) |
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Hard to read when the pages are stuck together.
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Sun King



Registered: 02/15/14
Posts: 4,069
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You should be more respectful of your library books.
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night_shift
Stranger

Registered: 01/02/15
Posts: 226
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Re: mathematical brain teaser thread *DELETED* [Re: Sun King] 2
#21959308 - 07/18/15 04:54 AM (8 years, 6 months ago) |
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Post deleted by night_shift
Reason for deletion: [delete]
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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Re: mathematical brain teaser thread [Re: night_shift]
#21959468 - 07/18/15 06:58 AM (8 years, 6 months ago) |
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Quote:
night_shift said:
Quote:
Sun King said:
Quote:
DividedQuantum said: PROBLEM #16:
Quote:
Ask spectator A to jot down any three-digit number, and then to repeat the digits in the same order to make a six-digit number (e.g. 394,394). Without showing you the number, A is then to pass the sheet of paper to spectator B, who is requested to divide the number by 7.
"Don't worry about the remainder, " you tell B, "because there won't be any." B is surprised to discover that you are right. Without telling you the result, B passes the number to spectator C, who is told to divide it by 11. Once again, you state that there will be no remainder, and this also proves correct.
Direct spectator D to divide the last result by 13. Again, the division comes out even. This final result is written on a slip of paper that is folded and handed to you. Without opening it you pass it on to A.
"Open this," you tell A, "and you will find your original three-digit number."
Prove that the trick cannot fail to work regardless of the original digits chosen.
It's witchcraft. I don't know how to prove witchcraft, other than writing a computer program and I'm too lazy for that.
I'll just trust you that it works.

The notion that witchcraft is somehow supernatural and/or not scientific is false. Here's the proof.
abcabc = 100000*a + 10000*b + 1000*c + 100*a + 10*b + c = = 1001*100*a + 1001*10*b + 1001*c = = 1001*(100*a + 10*b + c) = = 7*11*13*abc
Dancing naked under a full moon also helps, of course.
  
Correct! Beautiful!
Writing a three-digit number twice is the same as multiplying it by 1,001. This number has the factors 7, 11, and 13, so writing the chosen number twice is equivalent to multiplying it by 7, 11, and 13. Naturally, when the product is successively divided by these same three numbers, the final remainder will be the original number.
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extreme


Registered: 04/05/11
Posts: 9,340
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Quote:
DividedQuantum said: PROBLEM #13
Quote:
If you happen to meet two of the Jones sisters (this assumes that the two are random selections from the set of all the Jones sisters), it is an exactly even-money bet that both girls will be blue eyed. What is your best guess as to the total number of blue-eyed Jones sisters?
Quote:
secondorder said: Three?
Three blue eyed, and one not blue eyed.
You'd have a 75% chance of meeting a blue eyed Jones sister if you just met one of them. If you were to then meet another Jones sister, you'd have a 66% chance of meeting one of the remaining two blue eyed sisters.
Two thirds of three quarters is one half: 50% chance.
Sorry this post is kinda useless but I feel good about it so I'm gonna post. I think I'm pretty good at "mental math" and word problems and stuff. I was always exceptional at math going back to elementary school.
Anyway, problem #13 was the most recent post when I clicked on this thread. I read it and took an educated "guess" of 75% within seconds. Now I don't exactly think this was difficult, but I'm happy to see I still kinda know what I'm doing 
Carry on.
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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PROBLEM #17:
Quote:
When Professor Stanislaw Slapenarski, the Polish mathematician, walked down the down-moving escalator, he reached the bottom after taking 50 steps. As an experiment, he then ran up the same escalator, one step at a time, reaching the top after taking 125 steps. Assuming that the professor went up five times as fast as he went down (that is, took five steps to every one step before), and that he made each trip at a constant speed, how many steps would be visible if the escalator stopped running?
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extreme


Registered: 04/05/11
Posts: 9,340
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"Educated quick guess" - 60. How close am I? Or is it more in the middle of the 2 #s?
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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Re: mathematical brain teaser thread [Re: extreme]
#21966838 - 07/19/15 07:34 PM (8 years, 6 months ago) |
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Wrong and not acceptable thread behavior.
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extreme


Registered: 04/05/11
Posts: 9,340
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Sorry lol. Am I allowed to guess again? Or should I not even be "guessing?"
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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Re: mathematical brain teaser thread [Re: extreme]
#21966870 - 07/19/15 07:39 PM (8 years, 6 months ago) |
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There was something of a row about this a few posts ago. Guess and check will be accepted if you show all work, i.e. exactly how you calculated the answer. Otherwise, please show your math.
No overt guessing, if you please.
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