Magic !
A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience selects two of the three boxes, chooses one card from each and announces the sum of the numbers on the chosen cards. Given this sum, the magician identifies the box from which no card has been chosen.
How many ways are there to put all the cards into the boxes so that this trick always works? (Two ways are considered different if at least one card is put into a different box.)
How many ways are there to put all the cards into the boxes so that this trick always works? (Two ways are considered different if at least one card is put into a different box.)
Labels: mathschallenge
posted by Rajesh Lal at
10:38 AM
4 Comments:
The only way I can think up is to deal e.g.
1 to A, 2 to B, 3 to C then 4 to A and so on.
Then after dividing the value of a card by 3, all the
cards in A have remainder 1, all the cards in B have
remainder 2 and all the cards in C have remainder 0.
So the remainder of the sum of a pair taken randomly from:
A+B = 0 => C wasn't used,
B+C = 2 => A wasn't used,
A+C = 1 => B wasn't used.
Other than the 6 trivial permutations due to the 3 choices
of dealing to A, B or C first, and the two choices of the
deal direction, I don't think there is another solution.
umm.. which box has 34 cards instead of 33? since 100 is not evenly dividable by 3....
put the 100 in one box. 1 in another box. and the rest of the cards in the last box.
so thats 1 way.
put 1 in one box. and 99 and 100 in another box. and the rest of the cards in the last box.
so that the 2nd way.
put 1 and 2 in one box. 100 in another box and the rest in the last box.
3rd way.
i'm sure there could be a few more ways but thats it for me
Anonymous 1- the 100 goes into A (as it has
remainder 1 after dividing by 3.)
Anonymous 2 - nice trick with 1 and 100. The other
two suggestions don't work when the sum is 101.
1+100 or 2+99 are the possibilities in both cases.
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