Anaira Mitch wrote:If a, b, x and y are positive integers, what is the value of a - b?
(1) x^a= (x^b)+(x^b)+(x^b)
(2) y^a= (y^b)+(y^b)+(y^b)+(y^b)
Please help with above problem.
We know that a, b, x, and y are positive integers.
We have to find out the value of (a - b).
S1: x^a= (x^b)+(x^b)+(x^b)
x^a= 3.x^b
=> x^a/x^b = 3
=> x^(a-b) = 3
Keep in mind that x, a, and b are positive integers, and (a-b) need not be a positive integer. It can be positive, negative or 0 but cannot be a fraction.
x cannot be 1 since 1^(a-b) = 1 ≠3
x cannot be 2 since if 2^(a-b), to make it equal to 3, (a-b) must be between 1 and 2 (not an integer), which is not possible.
x can be 3 since 3^(a-b) = 3 if (a-b) = 1. We get (a-b)=1
x cannot be greater than 3, say 4, since if 4^(a-b), to make it equal to 3, (a-b) must be between 0 and 1 (not an integer), which is not possible.
So only possible values are: x = 3 and (a-b) = 1. Sufficient.
S2: y^a= (y^b)+(y^b)+(y^b)+(y^b)
y^a= 4.y^b
=> y^a/y^b = 4
=> y^(a-b) = 4
Keep in mind that y, a, and b are positive integers, and (a-b) need not be a positive integer. It can be positive, negative or 0 but cannot be a fraction.
Leveraging from the analysis done in S1, we get y=4 and (a-b) = 1
However, y^(a-b) = 4 can be written as y^(a-b) = 2^2
Thus, y=2 and (a-b)=2. No unique value of (a-b). Not sufficient.
Answer:
A
Hope this helps!
-Jay
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