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Question

The number of proper divisors of 2 to the power of p. 6 to the power of q. 15r is-

  1. (p + q + 1) (q + r + 1) (r + 1)    
  2. (p + q + 1) (q + r + 1) (r + 1) – 2    
  3. (p + q) (q + r) r – 2    
  4. None of these    

Hint:

A proper divisor of a natural number is the divisor that is strictly less than the number.
For example, number 20 has 5 proper divisors: 1, 2, 4, 5, 10, 20
Proper divisors of number 20 are 2,4,5 and 10 excluding 1 and 20(the number itself)

The correct answer is: (p + q + 1) (q + r + 1) (r + 1) – 2


      2 to the power of p6 to the power of q. 15 to the power of r  - We need to find proper divisors.
    Suppose a to the power of n is a number then factors of a to the power of n  = (a to the power of 0 comma space a to the power of 1 comma space a squared comma a cubed comma........ space a to the power of n right parenthesis
 and a is proper
    i.e. a to the power of n has total division = (n + 1)
    Now,   2 to the power of pcross times 6 to the power of qcross times 15 to the power of r = 2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r
    We know that x to the power of m cross times space x to the power of n space equals x to the power of m space plus space n end exponent
    Thus,   2 to the power of pcross times 6 to the power of qcross times 15 to the power of r = 2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r = 2 to the power of left parenthesis p plus q right parenthesis end exponent cross times 3 to the power of left parenthesis q plus r right parenthesis end exponent cross times 5 to the power of r
    Total factors = (p+q+1)(q+r+1)(r+1)

    However, proper divisors exclude 1 and the number itself.
    Hence, the answer is (p+q+1)(q+r+1)(r+1)2.

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