Question 33

The number of integers x such that $$0.25 \leq 2^x \leq 200$$ and $$2^x + 2$$ is perfectly divisible by either 3 or 4, is

** Correct Answer:** 5

Solution

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.Â

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.Â

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.Â

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.Â

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.Â

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.Â

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.Â

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.Â

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.Â

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