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Divisors Multiple Probability

XAT 2014 · Quantitative Ability & Data Interpretation · Hard · Probability

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The probability that a randomly chosen positive divisor of 102910^{29} is an integer multiple of 102310^{23} is a2/b2a^2 / b^2, then 'b - a' would be:

  1. A.

    8

  2. B.

    15

  3. C.

    21

  4. D.

    23

  5. E.

    45

Answer

D

Explanation

1029=229×52910^{29} = 2^{29} \times 5^{29}. Total divisors =(29+1)(29+1)=900= (29+1)(29+1) = 900. Multiple of 1023=223×523  ⟹  1029/1023=106=26×5610^{23} = 2^{23} \times 5^{23} \implies 10^{29} / 10^{23} = 10^6 = 2^6 \times 5^6. Divisors that are multiples =(6+1)(6+1)=49= (6+1)(6+1) = 49. Probability =49/900=(7/30)2  ⟹  a=7,b=30= 49/900 = (7/30)^2 \implies a=7, b=30. b−a=30−7=23b - a = 30 - 7 = 23.

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