(x·2)+x(2^2)+(y·2^3)+(y·2^4)=42
2x + 4x + 8y + 16y = 42
6x + 24y = 42
x + 4y = 7
x = 7 - 4y
so you are told that both x and y are POSITIVE integers,
what are the only possible values of y for x to be a positive integers
Given that x and y are positive integers and (x·2)+x(2^2)+(y·2^3)+(y·2^4)=42, what is the value of xy?
2 answers
x=3, y=1
So the answer is 3
:)
So the answer is 3
:)