All Exams Test series for 1 year @ ₹349 only
Question

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.

Was this answer helpful?

Similar Questions

  1. What is the sum of the largest and the smallest 4-digit numbers made by using single digit prime numbers (without repetition)?

  2. XYZ is a 3-digit number, where X, Y, Z are distinct non-zero digits. The difference between the two 3-digit numbers XYZ and YXZ is 90. How many possible values exist for the sum (X+Y) ?
  3. The number 199 can be written as \(m^2 - n^2\), where \(m, n\) are natural numbers (\(m > n\)). What is the value of \(mn\)?
  4. How many numbers of the form \(2^n - 1\) and less than 2000 are prime?
  5. If \(\frac{1}{a + \frac{1}{b + \frac{1}{c + \frac{1}{d + \frac{1}{e}}}}} = \frac{421}{972}\), then what is the value of \(a \times b \times c \times d \times e\)?
  6. Let \(p\) and \(q\) be natural numbers such that \(q > p\). What is the largest value of \(p\) such that \(q^2 - 5p-4\) is negative?
  7. Let \(x\) and \(y\) be natural numbers, each less than 20, such that \(x\), \(y\), \(x+y\) and \(x-y\) are prime numbers. How many such combinations of \((x, y, x+y, x - y)\) are possible?
  8. What is the remainder when 

    \((17^{25} +19^{25})\) 

    is divided by 18?

  9. Let \(x = n(n+1)(n+2)\), where \(n\) is an even natural number. Which of the following statements is/are correct? 

    I. \(x\) is always divisible by 48. 

    II. \(x^2\) is always divisible by 144. 

    Select the answer using the code given below.

  10. Let XYZ be a 3-digit number. Let D be the difference between XYZ and ZYX. What is the remainder when D is divided by 99?

Important Questions from Number System

  1. The number 253118139 is divisible by which of the following?
  2. The number of prime numbers lying between 476 and 485, both included, is:
  3. The product of two consecutive natural numbers is 6. The greater of the two numbers is:
  4. Which of the following numbers divides 386937355?
  5. If 1 is added to each odd digit and 2 is subtracted from each even digit in the number 8317456, what will be the sum of the digits that are second from the left and second from the right?

Need Expert Advice?
Test Series
CDS img
Defence
UPSC CDS 2026 Mock Test Series
559 Tests 4 Tests Free
1821 Attempts
4.3(195)
English, Hindi
More Questions from CDS

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App