The question asks for the minimum possible value for the sum \((p+q)\), where \(p\) and \(q\) are natural numbers (meaning \(p \ge 1\) and \(q \ge 1\)). The condition is that the expression \((p+q)^{p+q}\) must be perfectly divisible by 512.
First, let's find the prime factorization of 512.
\(512 = 2 \times 256 = 2 \times 2^8 = 2^9\)
So, the condition is that \((p+q)^{p+q}\) must be divisible by \(2^9\).
Let \(N = p+q\). Since \(p\) and \(q\) are natural numbers, the smallest possible value for \(N\) is \(1+1 = 2\). The condition becomes \(N^N\) must be divisible by \(2^9\).
For \(N^N\) to be divisible by \(2^9\), the prime factorization of \(N\) must include the prime factor 2. This means \(N\) must be an even number.
Let the prime factorization of \(N\) be \(N = 2^k \cdot m\), where \(m\) is an odd integer and \(k \ge 1\) (since N is even). Then, the expression \(N^N\) can be written as:
\(N^N = (2^k \cdot m)^N = (2^k)^N \cdot m^N = 2^{k \cdot N} \cdot m^N\)
For \(N^N\) to be divisible by \(2^9\), the exponent of 2 in its prime factorization must be at least 9. Therefore, we need:
\(k \cdot N \ge 9\)
We need to find the smallest integer \(N \ge 2\) such that \(N\) is even, and if \(N = 2^k \cdot m\) (with \(m\) odd), then \(k \cdot N \ge 9\). Let's test values:
Since we are looking for the least value of \((p+q)\), and \(N=8\) is the first value that satisfies the condition \(k \cdot N \ge 9\), it is the minimum possible value for \((p+q)\). We can easily find natural numbers \(p\) and \(q\) that sum to 8 (e.g., \(p=4, q=4\) or \(p=1, q=7\)).
The least value of \((p+q)\) such that \((p+q)^{p+q}\) is divisible by 512 is 8.
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Select the answer using the code given below.
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