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What is the probability that divisor of $10^{99}$ is a multiple of $10^{96}$?

  1. $\left(\dfrac{1}{625}\right)$
  2. $\left(\dfrac{4}{625}\right)$
  3. $\left(\dfrac{12}{625}\right)$
  4. $\left(\dfrac{16}{625}\right)$
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@Arjun Sir,Wouldn't the powers of 2 be 20,21,22,23 instead of 24. Similarly for 5 d powers be 50,51,52,53 instead of 5.Tat is how 4 options for 2 and 5 respectively. So for Numerator part 4X4 respectively. Similarly for 299 and 599. The denominator wud give 100 X 100. Hence 16/10000=1/625

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multiple of $2^{96}$

means 1st term is $2^{96}$, then $2^{96}\times 2$,$2^{96}\times 3$..... like that it will go
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2 Answers

107 votes
107 votes
Best answer

Prime factorization of $10 = 2 \times 5$.
So, $10^{99} = 2^{99} \times 5^{99}$ and

No. of possible factors for $10^{99} = $ No. of ways in which prime factors can be combined
$= 100 \times 100$ (1 extra possibility for each prime number as prime factor raised to 0 is also possible for a factor)

$10^{99} = 10^{96} \times 1000$
So, no. of multiples of $10^{96}$ which divides $10^{99} =$ No. of possible factors of 1000

$= 4\times4 \left( \because 1000 = 2^3 \times 5^3\right)$ (See below)

$=16$

So, required probability $= \frac{16}{10000}$

$= \frac{1}{625}$

Correct Answer: $A$


How is number of possible factors of $1000 = 16?$

Here, we can prime factorize $1000$ as $2^3 \times 5^3$. Now, any factor of $1000$ will be some combination of these prime factors. For $2$, a factor has $4$ options - $2^0, 2^1, 2^2$ or $2^3$. Similarly $4$ options for $5$ also. This is true for any number $n$, if $n$ can be prime factorized as $a_1^{m_1} . a_2^{m_2} \dots a_n^{m_n}$, number of factors of $n$ $\\ = \left(m_1 +1 \right) \times \left(m_2 + 1\right) \times \dots \times \left(m_n +1 \right) $, the extra one in each factor term coming for power being $0.$ 

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4 Comments

Nice explanation , got it!
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In the part where, required probability = 16/10000.

How 10^4 is there?
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@M_eight We know that if $n = a^{p}b^{q}$ then #factors of $n = (p+1)*(q+1)$. 

Similarly $10^{99} = (5*2)^{99} = 5^{99} *2^{99}$ number of factors of $10^{99}$ is $100*100 = 10,000$. Now out of these possible factors, how many are the multiples of $10^{96}$?

$10^{99}=10^{96}×1000$
So, no. of multiples of $10^{96}$ which divides $10^{99}$= No. of possible factors of 1000

=4×4($∵1000=2^3×5^3$)

=16

as explained by @Arjun sir. Therefore required probability is $\frac{16}{10,000}$

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24 votes
24 votes

Number of divisors of 10n=(n+1)2

Number of divisors of 1099=10,000

Number of divisors of 1099 which are multiples of 1096

Number of divisors of 103 =(3+1)2=16

Required probability =16/10000=1/625

Source: https://www.aspirebuzz.com/content/what-probability-divisor-1099-0

1 comment

Better and easier than above
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Answer:

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