Question

The number of positive integers $n$ in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\] such that $\dfrac{1}{n}$ has a terminating decimal expansion is:
a.16
b.18
c.40
d.100

Answer 1

Hint: Here, for each positive integer $n$ in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\], consider $\dfrac{1}{n}$ ,first we have to reduce $n$ into its prime factors, and if $n$ can be written in the form,$n={{2}^{m}}{{5}^{n}}$ where $m$ and $n$ are integers, then we can say that $\dfrac{1}{n}$ has a terminating decimal expansion. Then find how many such numbers are there in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\].

Complete step-by-step answer:
Here, we are given the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\].
Now we have to find the number positive integers $n$ in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\]such that $\dfrac{1}{n}$ has a terminating decimal expansion.
For each positive integer $n$ in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\], consider $\dfrac{1}{n}$ ,first we have to reduce $n$ into its prime factors, and if $n$ can be written in the form, $n={{2}^{m}}{{5}^{n}}$ where $m$ and $n$ are integers, then we can say that $\dfrac{1}{n}$ has a terminating decimal expansion.
That is, if the denominator $n$ contains only the factors of 2 and 5, then the decimal expression terminates. If there is any prime factor in the denominator other than 2 or 5 then the decimal expression repeats.
So, here in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\] we have to find the numbers that have prime factors only 2 and 5.
The numbers that have only 2 as the prime factor are $2,4,8,16,32,64,128$.
Therefore, we can say that there are 7 numbers in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\] that have only 2 as the prime factor.
Next, consider the numbers that have only 5 as the prime factor. They are $5,25,125$.
Hence we can say that there are only 3 numbers in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\] that have only 5 as the prime factor.
Next, we have to consider the numbers that have only 2 and 5 as the prime factors. They are $10,20,40,50,80,100,160,200$.
Hence, we can say that there are 8 numbers that have only prime factors 2 and 5.
Therefore, the total numbers that have the prime factors 2 or 5 or 2 and 5 = 7 + 3 + 8 = 18.
Hence, there are 18 positive numbers in the set \[\left\{ 2,\text{ }3,\text{ }4,\ldots ..,200 \right\}\] such that $\dfrac{1}{n}$ has a decimal expansion.
So, the correct answer for this question is option (b).

Note: Here, you have to consider only such $n$ in the set that has the prime factors 2 and 5. If any other prime factors such as 3, 5, 7, 11 etc occurs then the $\dfrac{1}{n}$, won’t have a terminating expansion. If you consider any number that has the prime factor other than 2 and 5, then you would get a wrong answer.