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NCERT Exemplar · Q3

Q.If Y={x∣x is a positive factor of the number 2p−1(2p−1), where 2p−1 is a prime number}Y = \{x \mid x \text{ is a positive factor of the number } 2^{p-1}(2^p - 1),\ \text{where } 2^p - 1 \text{ is a prime number}\}. Write YY in the roster form.

Rajasthan RbseShort· 2mImportance★★★★★est
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The set YY consists of all positive divisors of a perfect number of the form 2p−1(2p−1)2^{p-1}(2^p - 1) where 2p−12^p - 1 is prime. Since 2p−12^p - 1 is prime, the divisors are 1,2,4,…,2p−11, 2, 4, \dots, 2^{p-1} and each multiplied by the prime 2p−12^p - 1. So Y={1,2,22,…,2p−1,(2p−1),2(2p−1),22(2p−1),…,2p−1(2p−1)}Y = \{1, 2, 2^2, \dots, 2^{p-1}, (2^p - 1), 2(2^p - 1), 2^2(2^p - 1), \dots, 2^{p-1}(2^p - 1)\}.

The problem gives you a number of the form N=2p−1(2p−1)N = 2^{p-1}(2^p - 1), with the condition that 2p−12^p - 1 is prime. This is the classic form of an even perfect number — every even perfect number is of this shape, and conversely, whenever 2p−12^p - 1 is prime (a Mersenne prime), NN is perfect.

But we don't need the perfection property here. What matters is the factor structure. Since 2p−12^p - 1 is prime, call it qq. Then N=2p−1⋅qN = 2^{p-1} \cdot q, where qq is an odd prime and 2p−12^{p-1} is a power of 2. The two parts are coprime (one is a power of 2, the other is an odd prime), so the divisors of NN are simply all possible products of a divisor of 2p−12^{p-1} and a divisor of qq.

Let's list them systematically.

  1. Divisors of 2p−12^{p-1}: These are 1,2,22,23,…,2p−11, 2, 2^2, 2^3, \dots, 2^{p-1}. That's pp numbers.

  2. Divisors of qq (where q=2p−1q = 2^p - 1 is prime): Only 11 and qq itself.

  3. All divisors of NN: Take each divisor of 2p−12^{p-1} and multiply it by each divisor of qq. That gives:

    • Multiply by 11: 1,2,22,…,2p−11, 2, 2^2, \dots, 2^{p-1}
    • Multiply by qq: q,2q,22q,…,2p−1qq, 2q, 2^2 q, \dots, 2^{p-1} q

    No other combinations exist because qq has no other factors.

  4. Total count: There are pp divisors from the first row and pp from the second, so 2p2p divisors in all. This matches the divisor-count formula: if N=2p−1⋅q1N = 2^{p-1} \cdot q^1, then τ(N)=(p−1+1)(1+1)=p⋅2=2p\tau(N) = (p-1+1)(1+1) = p \cdot 2 = 2p.

Watch out

A common mistake is to forget that 11 is a divisor, or to think that 2p−12^p - 1 itself might factor further. The problem explicitly states 2p−12^p - 1 is prime, so it has exactly two divisors: 11 and itself.

Tip

Notice that the divisors come in natural pairs: each divisor dd from the first row pairs with N/dN/d from the second row. For example, 11 pairs with 2p−1q2^{p-1}q, 22 pairs with 2p−2q2^{p-2}q, and so on. This is a hallmark of perfect numbers — the sum of all divisors equals 2N2N.

So the roster form of YY is simply the list of all these 2p2p numbers.

✓Final answer

Y={1,2,22,…,2p−1,(2p−1),2(2p−1),22(2p−1),…,2p−1(2p−1)}Y = \{1, 2, 2^2, \dots, 2^{p-1}, (2^p - 1), 2(2^p - 1), 2^2(2^p - 1), \dots, 2^{p-1}(2^p - 1)\}

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