Powers of 6
Author: mathforces
Problem has been solved: 40 times
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Find the sum of all possible values of $ xy $, where $ x \geq 0 $ and $ y $ are integers and $ (6 ^ x-y) ^ 2 = 6 ^ {x + 1} -y $.
Найдите сумму всевозможных значений $xy$, где $x \geq 0$ и $y$ целые числа и $(6^x-y)^2=6^{x+1}-y$.
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