800+ problem

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800+ problem

by lime777 » Mon Feb 04, 2013 10:41 pm
How many pairs of positive integers x and y satisfy the equation x^y = 2013^2013 ? Explain

How many pairs of positive integers x and y satisfy the equation x^y = 2012^2012 ? Explain

This one is out of this rigorous prep course, no answer keys yet.
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by Anurag@Gurome » Tue Feb 05, 2013 12:04 am
lime777 wrote:How many pairs of positive integers x and y satisfy the equation x^y = 2013^2013?
2013 = 3*11*61

Hence, 2013^2013 = (3*11*61)^(3*11*61) = [(3*11*61)^3]^(11*61) = [(3*11*61)^11]^(3*61) = ... = [(3*11*61)^(3*11)]^61 = ... etc

Each of these above expression will give one pair of x and y. For example, for [(3*11*61)^11]^(3*61), x = [(3*11*61)^11] and y = (3*61)

Hence, number of pairs of x and y = Number of ways to select 1 or 2 or 3 integers among 3, 11, and 61 = (Number of ways to select 1 integer among 3 + Number of ways to select 2 integers among 3 + Number of ways to select 3 integers among 3) = 3C1 + 3C2 + 3C3 = 3 + 6 + 1 = 10
lime777 wrote:How many pairs of positive integers x and y satisfy the equation x^y = 2012^2012?
2012 = (2^2)*503
The situation similar to the previous one, but now we have two identical factors. Hence, number of pairs of x and y will be exactly half of the previous one, i.e. 10/2 = 5
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