Terminating Decimals ...

II · Posted: Sat Jan 05, 2008 12:41 pm Post subject: Terminating Decimals ...

Which of the following is a terminating decimal ?

A) 6/7

B) 5/33

C) 4/75

D) 3/128

E) 3/312

Can you please explain how you came to the solution ?

Thanks in advance.
II

gmatguy16 · Posted: Tue Jan 08, 2008 3:25 pm Post subject:

someone else had already discussed on the same forum that if the denominator can be expressed in terms of 2 ^ a + 5 ^b then the fraction is terminating ..so only 3/128 is terminating since 128 = 2 ^ 7

StarDust845 · Posted: Wed Jan 09, 2008 8:25 am Post subject:

parore26 · Posted: Wed Jan 09, 2008 9:50 am Post subject:

5^b for any integer b is never equal to 0. There are two ways to solve it.

The first way: solve for the reciprocal for each of the denominators and see if that is a terminating decimal. So in this case we woul'd solve for 1/7, 1/33, 1/75, 1/128, and 1/312.

You can easily eliminate A) & B) A quick division also eliminates c) the real tough one here is between d) & e) . You can solve for D and notice that it is terminating.

The second way: 1/(2^a) for a > 1 is terminating. Why?
multiply 1/(2^a) by (10^a)/(10^a) you get (5^a)/(10^a) . The trick here is to note that 5^a is a finite string of numbers that terminate. Therefore, by dividing by 10^a, you just shift the decimal point and the number is terminating.

gmatguy16 · Posted: Wed Jan 09, 2008 3:15 pm Post subject:

i think the rule to check for denominator is 2 ^ a + 5 ^ b still holds true,it means either a or b are zero ,both cannot be non-zero...

StarDust845 · Posted: Wed Jan 09, 2008 3:18 pm Post subject:

parore26 · Posted: Wed Jan 09, 2008 3:28 pm Post subject:

gmatguy,
I'm not sure i understand your point. Consider 1/3, it's not terminating and denominator 3 can be written as 3 = 2 + 1 = (2^1)+(5^0) The denominator now is of the form 2^a + 5^b yet 1/3 is a non-terminating decimal.

In your original post you suggested that 128 = 2^7 + 5^0 which is false. 2^7 + 5^0 = 129.

Any number to the power 0 is 1.

Stuart Kovinsky · Posted: Wed Jan 09, 2008 5:14 pm Post subject:

I think the rule is that a fraction will result in a terminating decimal if the denominator can be written in the form 2^a OR 5^b OR (2^a)(5^b) (for all integers a and b).

simplyjat · Posted: Thu Jan 10, 2008 12:01 am Post subject:

2 and 5 are the only prime numbers that produce non-repeating reciprocals.

So to find if any fraction will result in terminating or non-terminating decimal.
1. Reduce the fraction to the lowest terms.
2. List all the prime factors of denominator.
3. If denominator contains a prime factor other than 2 or 5 then the resulting fraction will be non-terminating.

Example.
1/256 = 1/( 2^8 ) -> terminating
7/625 = 1/( 5^4 ) -> terminating
9/30 = 3/10 = 1/( 2^1*5^1 ) -> terminating
1/75 = 1/( 3^1*5^2 ) -> non-terminating
9/75 = 3/25 = 3/( 5^2 ) -> terminating