Three-Digit Number Puzzle: Solving a Parlour Game with Modular Arithmetic

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The discussion revolves around solving a parlour game involving a three-digit number and its permutations. Participants are tasked with finding the original number after summing its permutations, which equals 3194. The key equation derived is 122a + 212b + 221c = 3194, leading to the insight that the sum of the digits (a+b+c) can help find the original number. By reducing the equation modulo 222, the possible values for the original number are narrowed down to four candidates: 136, 358, 580, and 802. Ultimately, the correct solution is identified as 358, demonstrating the effectiveness of modular arithmetic in solving the puzzle.
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came across this in a book I am reading, it doesn't give the answer though youll know once you get there that you're right anyways.

Homework Statement



In a parlour game, the 'magician' asks one of the participants to think of a three-digit number abc_{10}. Then the magician asks the participant to add the five numbers acb_{10},/bac_{10},/bca_{10},/cab_{10},/cba_{10}, and reveal their sum. Supppose the sum was 3194. What was the abc_{10} originally?

Homework Equations



none known

The Attempt at a Solution



right, I am not positive I'm doing the right thing, but I hope I am on the right track, just say where I need to go if I'm not, or correct me, or do whatever, thnx

acb_{10} + bac_{10} + bca_{10} + cab_{10} + cba_{10} = 3194

expand that

100a + 10c + b + 100b + 10a + c + 100b + 10c + a + 100c + 10a + b + 100c + 10b + a = 3194

122a + 212b + 221c = 3194

now, from here I'm kinda stuck. I have a hunch I have to use modular arithmetic somewhere (only just roughly learned it so that why I kinda stuck on how to use it).

If I was to guess I'd do something like

a (mod/122) + b (mod/212) + c (mod/221) = 3194

but even with that guess id be stuck.

I 'could' try do a lot of guesswork and go from there but I'm guessing there's a more elegant way. thnx

hope you can help
 
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Note that if you add all six permutations together, you get 222(a+b+c) as your answer.
Now, (a+b+c) is the cross-sum of each permutation, and the trick most likely consists in finding this cross-sum by aid of the known number.

Thus, given that you have found this, multiply it by 222 and subtract the sum you were told to find the original number.

Or something along these lines..
 
soz, but what do you mean by 'cross-sum'?

thnx
 
He means 'cross sum'=a+b+c. So he has shown 222*(a+b+c)-3194=abc. On way to go is to reduce mod 222. So you get -86=abc(mod 222). Which gives you only 4 possibilities 136, 358, 580, 802. Then using your first equation you can pretty quickly figure 358 is the right one. But that seems like a laborious trick. It would be nicer if you could find a clever way to deduce a+b+c directly. Then you wouldn't have to guess.
 
ooo rite, i see what you did there, everything bar the bit where you took

222*(a+b+c)-3194=abc and reduced it mod 222 to -86=abc(mod 222)

but i see how you attained the 4 possibles, so could you quickly explain the bit i just mentioned? thnx
 
222*(a+b+c)=0 mod 222, right?
 
o rite yeh, soz i see it now,

thnx for all the help with it

now i knew that i had to play with mod 222 I started from scrathc and did it all myself and got it right :D yey go me

thnx
 

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