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Question 11 : The probability that a randomly chosen factor of 10^{19} is a multiple of 10^{15} is

- \\frac{1}{25})
- \\frac{1}{12})
- \\frac{1}{20})
- \\frac{1}{16})

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10^{19} = 2^{19} × 5^{19}

This is of the form 2^{x} × 5^{y}

x can take values from 0 to 19

y can take values from 0 to 19

x can take 20 vales and y can also take 20 values

Chosen number has 20 × 20 = 400 factors

Among that we need a number which is a multiple of 10^{15} = 2^{15} × 5^{15}

This is of the form 2^{a} × 5^{b}

We need a number that is not only a multiple of 2^{a} × 5^{b}but also a factor of 2^{19} × 5^{19}

Looking for numbers which are in form 2^{p} × 5^{q}

Since 2^{p} × 5^{q} is a factor of 2^{19} × 5^{19}, p and q should be less than or equal to 19

Since 2^{p} × 5^{q} is a multiple of 2^{15} × 5^{15}, p and q should be more than or equal to 15

p and q can take any of the 5 values 15, 16, 17, 18, 19.

So, the probability is \\frac{5 × 5}{20 × 20}) = \\frac{1}{16})

The question is **" The probability that a randomly chosen factor of 10 ^{19} is a multiple of 10^{15} is " **

Choice C is the correct answer.

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