How many integers n are there, where 1

Question

How many integers  n are there, where  1 <= n <= 100,  (206+9n^(2))/(2+3n) is a number that does not repeat decimals? (Repeating os do not count)

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Simplify the Expression: We need to find the number of integers n between 1 and 100 for which the expression (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 5^0, which is 1. 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 3. 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 <= 100). We can quickly calculate that 2^8 = 256, which is less than 303, and 2^9 = 512, which is more than 303. So we only need to check k=2 to k=8.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: - For k=2, (2^2 - 2) = 2, which is not divisible by 3. - For k=3, (2^3 - 2) = 6, which is divisible by 3. - For k=4, (2^4 - 2) = 14, which is not divisible by 3. - For k=5, (2^5 - 2) = 30, which is divisible by 3. - For k=6, (2^6 - 2) = 62, which is not divisible by 3. - For k=7, (2^7 - 2) = 126, which is divisible by 3. - For k=8, (2^8 - 2) = 254, 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: - For k=3, n = (2^3 - 2)/3 = 6/3 = 2. - For k=5, n = (2^5 - 2)/3 = 30/3 = 10. - For k=7, n = (2^7 - 2)/3 = 126/3 = 42. Each of these values of n is an integer between 1 and 100.We have found 3 values of n (2, 10, and 42) for which the expression (206+9n^2)/(2+3n) results in a non-repeating decimal. Therefore, there are 3 integers n between 1 and 100 that satisfy the given condition.

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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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)