The number of integers x such that 0 ≤ 2^x ≤ 200, and 2^x+2

There are two conditions
(A)\(0 ≤ 2^x ≤ 200\), here x can take values 0 to 7 as \(2^0=1 ..2^7=128\) and
(B) \(2^x+2\) .....so, the values are \(2^0+2...2^1+2..2^7+2\)

Divisible by 3-- X as even number will make \(2^x+2\) divisible by 2 due to the remainder property ( as \(2^{(even)}\) will leave a remainder of 1 when divided by 3 and when you add 2 to it, the term is divisible by 3) or you can check with pattern too.. \(2^1+2=4\)..No, \(2^2+2=6\)..Yes
So, x as 0, 2, 4, 6 are possible.

Divisible by 4-- All terms having x greater than 1 will make 2^x divisible by 4, so when you add 2 it, it will never be divisible by 4..
.for example \(2^3+2..... 2^3\) is divisible by 4, so \(2^3+2\) will NOT be divisible..
check for just 0 and 1...\(2^0+2=3.\)..NO AND \(2^1+2=4\)..YES
So x as 1 is possible

Total values - 0, 1, 2, 4, 6. Hence 5 is the answer

D