x, y > 0, integers
"Is
\frac{2x}{y} an integer?"
"Is
2x a multiple of (divisible by)
y?"
"Is
y a factor of
2x?"
Statement 1) This statements says that
\frac{2x + 2}{y} = \frac{2(x+1)}{y} is an integer. In other words,
2(x+1) is a multiple of
y.
In order for both
2x and
2(x+1) to be divisible by
y,
y must be either
1 (factor of
2 and both
x and
(x+1)) or
2 (factor of
2). Thus if
y equals
1 or
2, then
\frac{2x}{y} is an integer. If
y has any other value under this constraint,
\frac{2x}{y} is not an integer.
You can also break this statement into two expressions:
\frac{2x + 2}{y} = \frac{2x}{y} + \frac{2}{y}. If
\frac{2x}{y} + \frac{2}{y}=integer, then in order for
\frac{2x}{y} to be an integer,
\frac{2}{y} must also be an integer. Thus if
y equals
1 or
2, then
\frac{2x}{y} is an integer. If
y has any other value,
\frac{2x}{y} is not an integer.
Insufficient.
Statement 2) This statement says that
\frac{y}{x} is an integer. So
y = x*k, where
k is some integer.
In order for
\frac{2x}{y} = \frac{2*x}{k*x} to be an integer,
\frac{2}{k} must be an integer, ie
k must equal either
1 or
2. Thus if
y = 2x or
y = x,
2x will be a multiple of
y. With any other value of
k,
2x will not be a multiple of
y.
Insufficient.
Combined)
From statement 1,
\frac{2x}{y} is an integer when
y={1,2}, and not an integer when
y has another value.
From statement 2:
\frac{2x}{y} is an integer when
y=2x or
y=x, and not an integer when
y is a different multiple of
x than
2.
So under the constraints of both statements,
\frac{2x}{y} is an integer when
(x,y) =
(1,1),
(2,2), or
(1,2).
It only takes one example that follows the two constraints but is not one of these 3 possibilities to prove the combined statements insufficient. A couple examples are
(x,y) = (1,4), or
(2,6).
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