How many digits are there in a number($x$) that contains only $3$,$4$,$5$ & $6$ when the sum of digits of $x$ and $2x$ is 900?

Denote the number of 3s, 4s, 5s, 6s with $a,b,c,d$. Obviously:

$$3a+4b+5c+6d=900\tag{1}$$

Consider what happens when you double the number. Each digit 3 becomes 6 (ignore carryovers for a moment) and increases the sum of digits by 3. Each digit 4 becomes 8 and increases the sum of digits by 4. Each digit 5 becomes 0 but adds 1 as a carryover thus reducing the sum of digits by 4. And each digit 6 becomes 2 but adds 1 as a carryover thus reducing the sum of digits by 3. Because the sum of digits cannot change, increases and decreases must cancel out:

$$3a+4b-4c-3d=0\tag{2}$$

Subtract (2) from (1) and you get:

$$9c+9d=900$$

or:

$$c+d=100\tag{3}$$

Multiply (3) by 5 and subtract from (1). You get:

$$3a+4b+d=400\tag{4}$$

Equations (3) and (4) are sufficient to find the minimum and maximum value of $x$.

Minimum value: Minimum value should have the smallest number of digits. Start from (4) and make $d$, than $b$ as big as possible. This gives you:

$$a=3, \ b=73, \ c=1, \ d=99$$

The smallest number has 176 digits and looks like this:

$$x_{\min}=333\underbrace{44\dots44}_\text{73 digits}5\underbrace{66\dots66}_\text{99 digits}$$

Maximum value: Maximum value should have as many digits as possible. Start from (4) and maximize $a$. This gives you:

$$a=131, \ b=1, \ c=97, \ d=3$$

The biggest number has 232 digits and looks like this:

$$x_{\max}=666\underbrace{55\dots55}_\text{97 digits}4\underbrace{33\dots33}_\text{131 digits}$$

Notice that:

$$3\times 10^{175}\lt x_{\min}\lt 4\times 10^{175}$$

$$6\times 10^{231}\lt x_{\max}\lt 7\times 10^{231}$$

Therefore:

$$18\times 10^{406}\lt x_{\min} x_{\max}\lt 28\times 10^{406}$$

...so the product of minimum and maximum value of $x$ must have exactly 408 digits.