 For all $n \in N$, ${10^n} + {3.4^{n + 2}} + 5$ is divisible byA.23B.3C.9D.207 Verified
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Hint: In the given question, we have been given that there is an expression involving four constants. The constants raised to some variable power. We are asked when some value is put in the variable power, and the answer we get, by what number is it always divisible by. First, we need to look at the options here. There can be multiple answers depending on the options. If only the option A is correct, then only A is correct. If only B is correct, then only B is correct. If option C is correct, then C as well as B is correct (because B is a factor of C). And, if D is correct, then all options are correct, because each option is a factor of the D option. Now, to solve this question, we just need to put in some values in the given range and see what answer we get.

The given expression is $p\left( x \right) = {10^n} + {3.4^{n + 2}} + 5$ and its range is $N$.
So, we put in the smallest possible value $\left( 1 \right)$ and see what we get. Now, if we get any of the options, then it means that by putting in any other value, we are going to get the same result, as $1$ divides every other natural number.
So, $p\left( 1 \right) = {10^1} + {3.4^{1 + 2}} + 5 = 10 + 3 \times {4^3} + 5$
Now, we know ${4^3} = 64$, so
$p\left( 1 \right) = 10 + 192 + 15 = 207$
So, on the first attempt only we got the answer that each of the options is going to divide the value of the expression, as $207$ is a multiple of every other number – $3,9,23$.