What is the smallest positive integer $n$ such that $\frac{n}{n+101}$ is equal to a terminating decimal?

n = 24

Step-by-step explanation:

Given the fraction:

To find:

Smallest positive integer 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 , only then the fraction will be equal to a terminating decimal.

Now, let us have a look at the denominator,

Let us use hit and trial method to find the value of as positive integer.

n = 1, denominator becomes 102 = not of the form .

n = 4, denominator becomes 105 = not of the form .

n = 9, denominator becomes 110 = not of the form .

n = 14, denominator becomes 115 = not of the form .

n = 19, denominator becomes 120 = not of the form .

n = 24, denominator becomes 125 = It is of the form .

So, the answer is n = 24

step-by-step explanation:

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