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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

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 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 $2^4$. 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.

by Veteran (431k points)
edited
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=4×4(1000=23×53)

=16

So, required probability =1610000

=1625

this part not getting plz explain

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explained, you can see last part.
+2
The question says that the divisor of 10^99 should be a multiple of 10^96 which means that the divisor must be divisible by 10^96 , but in the link mentioned above ,in the first answer it says that for the number to be a multiple of 10^96 we need to take numbers from 2^96 , I just took an example , like 2^4 =16 and 10^4 =10000, Now here 10^4 is a multiple of 2^4 ,not the other way round , since 2^4 divides 10^4 ,so then how can 10^96 divide 2^96 ?
+1

For 2, a factor has 4 options - $2^{0},2^{1},2^{2} or\, 2^{4}$

@Arjun Sir, Is last one typo? Should be $2^{3}$?

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Nice explanation , got it!

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

by Active (4.7k points)
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Better and easier than above