CAT 2018 Slot 1 — QA Question 27

Given, 0.25 ≤ 2^x ≤ 200
When x = −2, 2^x = 0.25 and x = 7 is the largest value that x can take for 2^x ≤ 200
∴ x = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7}

Now let us figure out for which values of x is 2^x + 2 perfectly divisible by either 3 or 4

For x = −2 and −1, (2^x + 2) is not an integer.

For x = 0, (2^x + 2) = 3 i.e., divisible by 3.

For x = 1, (2^x + 2) = 4 i.e., divisible by 4.

For x = 2, (2^x + 2) = 6 i.e., divisible by 3.

For x = 3, (2^x + 2) = 10 i.e., neither divisible by 3 nor by 4.

For x = 4, (2^x + 2) = 18 i.e., divisible by 3.

For x = 5, (2^x + 2) = 34 i.e., neither divisible by 3 nor by 4.

For x = 6, (2^x + 2) = 66 i.e., divisible by 3.

For x = 7, (2^x + 2) = 130 i.e., neither divisible by 3 nor by 4.

Thus, for x = 0, 1, 2, 4 and 6, 2^x + 2 is perfectly divisible by either 3 or 4.

Hence, 5.