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If (2^(200)-2^(192).31+2^n) is the perfe...

If `(2^(200)-2^(192).31+2^n)` is the perfect square of a natural number , then find the sum of digits of 'n' .

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To solve the problem, we need to determine the value of \( n \) such that the expression \( 2^{200} - 2^{192} \cdot 31 + 2^n \) is a perfect square of a natural number. We will follow these steps: ### Step 1: Factor out \( 2^{192} \) We start with the expression: \[ 2^{200} - 2^{192} \cdot 31 + 2^n \] We can factor out \( 2^{192} \): \[ 2^{192}(2^8 - 31 + 2^{n-192}) \] This simplifies to: \[ 2^{192}(256 - 31 + 2^{n-192}) = 2^{192}(225 + 2^{n-192}) \] ### Step 2: Set the inner expression as a perfect square For the entire expression to be a perfect square, the term \( 225 + 2^{n-192} \) must also be a perfect square. Let’s denote: \[ k = n - 192 \] So, we need \( 225 + 2^k \) to be a perfect square. Let’s denote this perfect square as \( m^2 \): \[ m^2 = 225 + 2^k \] ### Step 3: Rearranging the equation Rearranging gives us: \[ m^2 - 225 = 2^k \] This can be factored as: \[ (m - 15)(m + 15) = 2^k \] ### Step 4: Analyzing the factors Since \( m - 15 \) and \( m + 15 \) are two factors of \( 2^k \), both must be powers of 2. Let: \[ m - 15 = 2^a \quad \text{and} \quad m + 15 = 2^b \] where \( a < b \) and \( a + b = k \). ### Step 5: Finding the values of \( a \) and \( b \) From the equations, we have: \[ 2^b - 2^a = 30 \] Factoring gives: \[ 2^a(2^{b-a} - 1) = 30 \] The factors of 30 are \( 1, 2, 3, 5, 6, 10, 15, 30 \). We can test these to find valid \( a \) and \( b \). ### Step 6: Testing values 1. **If \( 2^a = 2 \) (i.e., \( a = 1 \))**: \[ 2^{b-a} - 1 = 15 \implies 2^{b-1} = 16 \implies b - 1 = 4 \implies b = 5 \] Thus, \( k = a + b = 1 + 5 = 6 \). 2. **If \( 2^a = 6 \)**, it is not a power of 2. Continuing this way, we find that \( a = 1 \) and \( b = 5 \) is the only valid solution. ### Step 7: Calculate \( n \) Since \( k = 6 \): \[ n - 192 = 6 \implies n = 198 \] ### Step 8: Find the sum of the digits of \( n \) Now we need to find the sum of the digits of \( n = 198 \): \[ 1 + 9 + 8 = 18 \] ### Final Answer The sum of the digits of \( n \) is \( \boxed{18} \).
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