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# Solution

Suppose Ashish gave me "M" cards and kept "A" cards with himself. Now, I have to transfer a certain number of my cards to him. Let me give him "x" cards. So, now I am left with (M - x) cards and Ashish will have (A + x) cards. In this situation, Ashish has four times as many cards as I have. So, (A + x) = 4(M - x). So, A + x = 4M - 4x; or A = 4M - 5x.

On the other hand, if Ashish transfers the same number, "x", of his cards to me, he will have (A - x) cards and I will have (M + x) cards. In this situation Ashish has thrice as many cards as I have. So, (A - x) = 3(M + x). Thus, A - x = 3M + 3x; or A = 3M + 4x.

Therefore, 4M - 5x = 3M + 4x, or M = 9x.

This means there are an infinite number of solutions because "x" can take any number of values and "M" will change accordingly. For instance, if x = 1 then M = 9; A = 3M + 4x = 27 + 4 = 31. Likewise, if x = 2, then M = 18 and A = 62; if x = 3, then M = 27 and A = 93; and so on ...

However, we are dealing with a standard deck of cards; in a standard deck there are 52 cards. So, (A + M) cannot exceed 52. If we consider x = 2, then M = 18 and A = 62 and (A + M) = 80, which is greater than 52. Thus, for any value of "x" equal to or greater than 2, (A + M) will be greater than 52. So, "x" has to be less than 2. Now, x cannot be zero because I must give Ashish a certain number of my cards. Thus, "x" has to be a number which is greater than zero and less than 2. But we are talking of cards, which cannot be dealt in fractions. So, "x" has to be a whole number. The only possibility, therefore, is that x = 1.

So, x = 1, M = 9, and A = 31. Thus, I have 9 cards, Ashish has 31 cards and the number of cards which is transferred to satisfy the conditions is 1. Let's do a quick check: If I give Ashish one of my cards I am left with 8 cards and Ashish has 32 which is four times eight. If on the other hand, Ashish gives me one of his cards I have 10 cards and Ashish has 30 cards which is three times ten.

Did I get it right?

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