  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

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.

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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