Answer:
n = 24
Step-by-step explanation:
Given the fraction:
[tex]$\frac{n}{n+101}$[/tex]
To find:
Smallest positive integer [tex]$n$[/tex] such that the fraction is equal to a terminating decimal.
Solution:
The rule that a fraction is equal to a terminating decimal states that, the denominator must contain factors of only 2 and 5.
i.e. Denominator must look like [tex]2^m\times 5^n[/tex], only then the fraction will be equal to a terminating decimal.
Now, let us have a look at the denominator, [tex]n+101[/tex]
Let us use hit and trial method to find the value of [tex]n[/tex] as positive integer.
n = 1, denominator becomes 102 = [tex]2 \times 3 \times 17[/tex] not of the form [tex]2^m\times 5^n[/tex].
n = 4, denominator becomes 105 = [tex]5 \times 3 \times 7[/tex] not of the form [tex]2^m\times 5^n[/tex].
n = 9, denominator becomes 110 = [tex]2 \times 5 \times 11[/tex] not of the form [tex]2^m\times 5^n[/tex].
n = 14, denominator becomes 115 = [tex]5 \times 23[/tex] not of the form [tex]2^m\times 5^n[/tex].
n = 19, denominator becomes 120 = [tex]5 \times 3 \times 2^3[/tex] not of the form [tex]2^m\times 5^n[/tex].
n = 24, denominator becomes 125 = [tex]2^0 \times 5 ^3[/tex] It is of the form [tex]2^m\times 5^n[/tex].
So, the answer is n = 24
