What is the largest possible number of cards I could have had in the box in the beginning?There are several cards in my box. A positive integer is wrritten on each of the cards so that he numbers...

What is the largest possible number of cards I could have had in the box in the beginning?

There are several cards in my box. A positive integer is wrritten on each of the cards so that he numbers on different cards are different. I take two cards from the box and replace them with a new card on which I write the product of the difference of the numbers on these two cards and their sum. For example, if the numbers on the two cards were 7 and 4, then the number on the new card would be (7-4)(7+4) = 33. After I repeat this procedure a few times, there is only one card left in the box with the number 2764 written on it. What is the largest possible number of cards I could have had in the box in the beginning?

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lfryerda | High School Teacher | (Level 2) Educator

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Consider at any given stage, where the cards are a and b.  Then the factors of the product are going to be `a+b` and `a-b` with product `a^2-b^2` .

Now, the final product is 2764 which has factor pairs 1 and 2764, 2 and 1382, 4 and 691.

If we call the larger factor u and the smaller factor l, then we can find the factors by solving the system

`a-b=u`

`a+b=l`

which means we can find the factors a and b from the pairs u and l to be `a={u+l}/2` and `b={u-l}/2` .

This means that for a and b to be integers, then both `u+l` and `u-l` must be even.  In this case, that is only the pair 2 and 1382, which means that `a=692` and `b=690` .

692 has factor pairs 1 and 692, 2 and 346, 4 and 173.  The only pair that is suitable is 2 and 346, which give the lower number of 172 and the upper number of 174.

The problem is that using 174 decomposes into factor pairs 1 and 174, 2 and 87, 3 and 58, 6 and 29.  None of these pairs can be used to generate new pairs since the are all even/odd combinations.

The number 172 has factors pairs 1 and 172, 2 and 86, 4 and 43.  The only pair that works is 2 and 86 which gives the new pairs 42 and 44.

The number 42 has factor pairs 1 and 42, 2 and 21, 6 and 7 which cannot generate new pairs.

The number 44 has factor pairs 1 and 44, 2 and 22, 4 and 11, so the only pair that works is 2 and 22, which generates the new pairs 10 and 12.

The number 10 only has factor pairs 1 and 10, 2 and 5, which cannot generate new pairs.

The number 12 has factor pairs 1 and 12, 2 and 6, 3 and 4.  The only number that can generate new pairs are 2 and 6 which make the pairs 2 and 4.

The numbers 2 and 4 cannot generate new pairs, since the numbers must be positive (0 is not positive).

In the other tree, we see that 690 has factor pairs 1 and 690, 2 and 345, 3 and 230, 5 and 138, 6 and 115, 10 and 69, 15 and 46, 23 and 30.  None of these combinations can be used to generate new pairs for the same reason.

This means that to get 2764, the only possible card combinations are {2764} or {690, 692} or {172, 174, 690} or  {42, 44, 174, 690} or {10, 12, 42, 44, 174, 690} or {2, 4, 10, 42, 44, 174, 690}.

The largest number of possible cards in the box at the beginning is 7.

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