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

We need to find integer values of $x$ that satisfy two conditions:

1. $0.25 < 2^x < 200$
2. $2^x + 2$ is divisible by either 3 or 4

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First, we'll convert 0.25 to a power of 2:

$0.25 = \frac{1}{4} = 2^{-2}$

So the inequality becomes:

$2^{-2} < 2^x < 200$

For the left side: $2^{-2} < 2^x$

Since the exponential function $2^x$ is increasing, this means:

$-2 < x$, so $x > -2$

For the right side: $2^x < 200$

Let us check some powers of 2:

• $2^7 = 128 < 200$
• $2^8 = 256 > 200$

So we need $x ≤ 7$ (since $x$ must be an integer)

Combining both conditions: $-2 < x ≤ 7$

Since $x$ must be an integer: $x \in \{-1, 0, 1, 2, 3, 4, 5, 6, 7\}$

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Now we'll calculate $2^x + 2$ for each possible value of $x$:

$x = -1$: $2^{-1} + 2 = 0.5 + 2 = 2.5$

Since this isn't an integer, it can't be divisible by 3 or 4.

$x = 0$: $2^0 + 2 = 1 + 2 = 3$

$3 ÷ 3 = 1$ (divisible by 3)

$x = 1$: $2^1 + 2 = 2 + 2 = 4$

$4 ÷ 4 = 1$ (divisible by 4)

$x = 2$: $2^2 + 2 = 4 + 2 = 6$

$6 ÷ 3 = 2$ (divisible by 3)

$x = 3$: $2^3 + 2 = 8 + 2 = 10$

$10 ÷ 3 = 3.33...$ and $10 ÷ 4 = 2.5$ (not divisible by 3 or 4)

$x = 4$: $2^4 + 2 = 16 + 2 = 18$

$18 ÷ 3 = 6$ (divisible by 3)

$x = 5$: $2^5 + 2 = 32 + 2 = 34$

$34 ÷ 3 = 11.33...$ and $34 ÷ 4 = 8.5$ (not divisible by 3 or 4)

$x = 6$: $2^6 + 2 = 64 + 2 = 66$

$66 ÷ 3 = 22$ (divisible by 3)

$x = 7$: $2^7 + 2 = 128 + 2 = 130$

$130 ÷ 3 = 43.33...$ and $130 ÷ 4 = 32.5$ (not divisible by 3 or 4)

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The values of $x$ that satisfy both conditions are: $x \in \{0, 1, 2, 4, 6\}$

The number of integers $x$ is 5.