What is the smallest positive integer \(n\) such that \(\frac{n}{2010}\) is a terminating decimal?

tertre
Dec 27, 2017

#2**+1 **

the prime factors of 2010 are 2,5,3 and 67

2010=2*5*3*67

In order for a fraction to be a terminating decimal you must first write it in its simplest from

THEN the prime factors of the denominator must be only 2 and 5.

so if n/2010 is to terminate then the 3 and 67 must be cancelled out when the fraction is simplified.

3*67=201.

So the smallest positive interger value of n is 201 if the decimal is to terminate.

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you do not really need to find the prime factors of 2010

You can just factor out the powers of 10 then the powers of 2 and 5 and see what is left.

2010 = 10*201

201 is not a multiple of 2 or 5 so 201 must be the smallest positive numerator if this fraction is going to terminate.

Melody
Dec 28, 2017

#2**+1 **

Best Answer

the prime factors of 2010 are 2,5,3 and 67

2010=2*5*3*67

In order for a fraction to be a terminating decimal you must first write it in its simplest from

THEN the prime factors of the denominator must be only 2 and 5.

so if n/2010 is to terminate then the 3 and 67 must be cancelled out when the fraction is simplified.

3*67=201.

So the smallest positive interger value of n is 201 if the decimal is to terminate.

--------------

you do not really need to find the prime factors of 2010

You can just factor out the powers of 10 then the powers of 2 and 5 and see what is left.

2010 = 10*201

201 is not a multiple of 2 or 5 so 201 must be the smallest positive numerator if this fraction is going to terminate.

Melody
Dec 28, 2017