Question 549955: Sum of three positive integers is 2^2011 + 1, and the product of two of them is 2^2011.how many values can the third integer take?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Since the product of two of these integers is a power of 2, it follows that both integers are powers of 2. Hence these integers can be {1, 2^2011}, {2, 2^2010}, ..., {2^1005, 2^1006} (order doesn't matter). The two integers cannot be {1, 2^2011} since the third integer would be zero. However, each of the other ordered pairs produces a unique integer (e.g. subtract the sum of the two integers from 2^2011 + 1), so the number of values for the third integer is the number of ordered pairs in {2, 2^2010}, ..., {2^1005, 2^1006}, or 1004 integers.
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