**Question: **

The number of integers x such that 0.25 < 2^{x} < 200, and 2^{x} +2 is perfectly divisible by either 3 or 4, is

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Correct Answer: 5

At $x = 0, 2^x = 1$ which is in the given range [0.25, 200]

$2^x + 2$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution.

At $x = 1, 2^x = 2$ which is in the given range [0.25, 200]

$2^x + 2$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution.

At $x = 2, 2^x = 4$ which is in the given range [0.25, 200]

$2^x + 2$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution.

At $x = 3, 2^x = 8$ which is in the given range [0.25, 200]

$2^x + 2$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can't be a solution.

At $x = 4, 2^x = 16$ which is in the given range [0.25, 200]

$2^x + 2$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution.

At $x = 5, 2^x = 32$ which is in the given range [0.25, 200]

$2^x + 2$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution.

At $x = 6, 2^x = 64$ which is in the given range [0.25, 200]

$2^x + 2$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution.

At $x = 7, 2^x = 128$ which is in the given range [0.25, 200]

$2^x + 2$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution.

At $x = 8, 2^x = 256$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7.

Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values.

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