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Let \(p\) and \(q\) be two natural numbers such that \((p+q)^{p+q}\) is divisible by 512. What is the least value of \((p+q)\)?

This question was previously asked in
CDS 2 2025 Maths Question Paper (14-Sep-2025)
The correct answer is
8

Understanding the Core Problem

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.

Analyzing the Divisibility Condition

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 \(2^9\).

Relating Divisibility to Prime Factors

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:

\(k \cdot N \ge 9\)

Finding the Least Value of N = (p+q)

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:

  • If \(N = 2\): \(N = 2^1 \cdot 1\). So, \(k=1\). \(k \cdot N = 1 \cdot 2 = 2\). Since \(2 < 9\), this doesn't work. (\(2^2 = 4\), not divisible by 512).
  • If \(N = 4\): \(N = 2^2 \cdot 1\). So, \(k=2\). \(k \cdot N = 2 \cdot 4 = 8\). Since \(8 < 9\), this doesn't work. (\(4^4 = 256\), not divisible by 512).
  • If \(N = 6\): \(N = 2^1 \cdot 3\). So, \(k=1\). \(k \cdot N = 1 \cdot 6 = 6\). Since \(6 < 9\), this doesn't work. (\(6^6 = 46656\), \(46656 / 512 \approx 91.1\), not divisible).
  • If \(N = 8\): \(N = 2^3 \cdot 1\). So, \(k=3\). \(k \cdot N = 3 \cdot 8 = 24\). Since \(24 \ge 9\), this condition is satisfied. (\(8^8 = (2^3)^8 = 2^{24}\), which is clearly divisible by \(2^9\)).

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 \(p=1, q=7\)).

Conclusion

The least value of \((p+q)\) such that \((p+q)^{p+q}\) is divisible by 512 is 8.

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