How many integers $n$ are there, where $1 \leq n \leq 100$ , $\newline$ $\frac{206+9 n^{2}}{2+3 n}$ $\newline$ is a number that does not repeat decimals? $\newline$ (Repeating os do not count)

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Q. How many integers

n

are there, where

1 \leq n \leq 100

\newline

\frac{206+9 n^{2}}{2+3 n}

\newline

is a number that does not repeat decimals?

\newline

(Repeating os do not count)

Simplify the Expression: We need to find the number of integers $n$ between $1$ and $100$ for which the expression $\frac{206+9n^2}{2+3n}$ results in a non-repeating decimal. A non-repeating decimal occurs when the denominator (after simplification) is a power of $2$ , a power of $5$ , or a product of powers of $2$ and $5$ , since any other prime factor would result in a repeating decimal. Let's start by simplifying the expression.
Factor Out Denominator: First, we notice that the denominator $2+3n$ can be factored out of the numerator $206+9n^2$ if we rewrite $206$ as $9n^2 - 9n^2 + 206$ . This gives us the expression $(9n^2 - 9n^2 + 206)/(2+3n)$ .
Split the Fraction: Now, we can split the fraction into two parts: $(9n^2)/(2+3n)$ and $(206-9n^2)/(2+3n)$ . The first part, $(9n^2)/(2+3n)$ , simplifies to $9n/2$ when $n$ is an integer, because $9n^2$ is divisible by $3n$ . The second part, $(206-9n^2)/(2+3n)$ , does not simplify in a similar way.
Analyze Denominator: We can ignore the first part $(9n/2)$ because it will always result in an integer when $n$ is an integer. We need to focus on the second part, $(206-9n^2)/(2+3n)$ , and determine when this expression results in a non-repeating decimal. To do this, we need to find when $2+3n$ is a power of $2$ , a power of $5$ , or a product of powers of $2$ and $5$ .
Set Up Equations: Since $2+3n$ must be a power of $2$ , a power of $5$ , or a product of powers of $2$ and $5$ , we can set up equations for each case and solve for $n$ . However, we notice that $2+3n$ cannot be a power of $5$ because when $n$ is an integer, $2+3n$ will always be odd, and powers of $5$ are always even except for $2$ $1$ , which is $2$ $2$ . Therefore, we only need to consider powers of $2$ .
Solve for $n$ : We set up the equation $2+3n = 2^k$ , where $k$ is a non-negative integer. We can then solve for $n$ in terms of $k$ . Subtracting $2$ from both sides gives us $3n = 2^k - 2$ . Dividing both sides by $3$ gives us $n = (2^k - 2)/3$ .
Test Values of $k$ : We need to find values of $k$ such that $n$ is an integer between $1$ and $100$ . Since $2^k$ is always even, $2^k - 2$ is also even, and thus divisible by $2$ . However, for $n$ to be an integer, $2^k - 2$ must also be divisible by $k$ $0$ . We need to find values of $k$ for which this is true.
Check Divisibility: We can test values of $k$ starting from $k=2$ (since $2^1 - 2 = 0$ , which does not give us a positive $n$ ) and going up to the point where $2^k$ exceeds $303$ (since $3n$ cannot exceed $303$ for $n \leq 100$ ). We can quickly calculate that $2^8 = 256$ , which is less than $303$ , and $k=2$ $1$ , which is more than $303$ . So we only need to check $k=2$ to $k=2$ $4$ .
Identify Valid n Values: For each value of $k$ from $2$ to $8$ , we check if $(2^k - 2)$ is divisible by $3$ . We find that: $\newline$ - For $k=2$ , $(2^2 - 2) = 2$ , which is not divisible by $3$ . $\newline$ - For $k=3$ , $(2^3 - 2) = 6$ , which is divisible by $3$ . $\newline$ - For $2$ $1$ , $2$ $2$ , which is not divisible by $3$ . $\newline$ - For $2$ $4$ , $2$ $5$ , which is divisible by $3$ . $\newline$ - For $2$ $7$ , $2$ $8$ , which is not divisible by $3$ . $\newline$ - For $8$ $0$ , $8$ $1$ , which is divisible by $3$ . $\newline$ - For $8$ $3$ , $8$ $4$ , which is not divisible by $3$ .
Final Conclusion: We have found that for $k=3, 5,$ and $7$ , $(2^k - 2)$ is divisible by $3$ . This gives us the corresponding values of $n$ as: $\newline$ - For $k=3$ , $n = (2^3 - 2)/3 = 6/3 = 2$ . $\newline$ - For $k=5$ , $n = (2^5 - 2)/3 = 30/3 = 10$ . $\newline$ - For $k=7$ , $7$ $0$ . $\newline$ Each of these values of $n$ is an integer between $7$ $2$ and $7$ $3$ .
Final Conclusion: We have found that for $k=3, 5,$ and $7$ , $(2^k - 2)$ is divisible by $3$ . This gives us the corresponding values of $n$ as: $\newline$ - For $k=3$ , $n = (2^3 - 2)/3 = 6/3 = 2$ . $\newline$ - For $k=5$ , $n = (2^5 - 2)/3 = 30/3 = 10$ . $\newline$ - For $k=7$ , $7$ $0$ . $\newline$ Each of these values of $n$ is an integer between $7$ $2$ and $7$ $3$ .We have found $3$ values of $n$ ( $7$ $6$ and $7$ $7$ ) for which the expression $7$ $8$ results in a non-repeating decimal. Therefore, there are $3$ integers $n$ between $7$ $2$ and $7$ $3$ that satisfy the given condition.