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Mr. X is a famous magician. He has 1 to 100 cards at his disposal. He puts them in 3 different boxes-red,green,blue. Now, he requests the audience to blindfold him and select 1 card each from any 2 different boxes and tell him the sum of the numbers obtained in those 2 cards. Say one card has 5 in it and the other 83, then the audience will say 88. Just by listening to the sum Mr. X points out the box that has not been selected.

In how many ways, can Mr. X distribute the cards in the 3 boxes such that each box has at least one card?

The answer given is 12, but I am getting 6.

My attempt

The only way the magician can know before hand if he uses the remainders of three. Say he groups in such a way that 1st box contains say multiples of three, 2nd box numbers which give remainder 1 when divided by 3, 3rd box remainder 2. Now when the audience adds the two numbers he/she will get only 3 combinations: 1+0=1; 1+2=0; 2+0=2. So, the magician can immediately identify the boxes. This he can do 3!=6 ways in 3 different boxes (permutation of 012). Hence the answer is 6.

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    $\begingroup$ Could you describe how you get $6$, and I can help you find your mistake? I have my suspicions, but I need to know your method to be sure. $\endgroup$ – Johanna Dec 30 '14 at 4:06
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    $\begingroup$ The only way the magician can know before hand if he uses the remainders of three. Say he groups in such a way that 1st box contains say multiples of three, 2nd box numbers which give remainder 1 when divided by 3, 3rd box remainder 2. Now when the audience adds the two numbers he/she will get only 3 combinations: 1+0=1;1+2=0;2+0=2. So, the magician can immediately identify the boxes. This he can do 3!=6 ways in 3 different boxes( Permutation of 012). Hence the answer is 6. $\endgroup$ – archangel89 Dec 30 '14 at 4:10
  • $\begingroup$ Total number of cards=100. $\endgroup$ – archangel89 Dec 30 '14 at 4:10
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    $\begingroup$ Total cards are 100. On the face of the 1st card is written 1, 2nd card 2 and so on and so forth. Now the trick happens. $\endgroup$ – archangel89 Dec 30 '14 at 4:14
  • $\begingroup$ I suspect you got downvotes because you didn't say how you got 6 or because you used irrelevant tags. Next time you should put your thoughts in your question itself, and it would make a good question. And you should try to use appropriate tags. $\endgroup$ – user21820 Dec 30 '14 at 4:50
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[1], [100], [2-99] gives the remaining 6 solutions

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  • $\begingroup$ Excellent. I don't know how i missed it. Thanks for answering.!!!!! $\endgroup$ – archangel89 Dec 30 '14 at 9:19
  • $\begingroup$ @archangel89 You have asked a total of $6$ questions on this SE and haven't accepted any of them. If you think Drew answered your questions clearly, you can accept his answer by clicking on the $\checkmark$ on the left side of his answer. You should perform this with other of your questions also. $\endgroup$ – Dheeraj Kumar Dec 31 '14 at 12:12

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