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
Terminating Decimals ...
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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
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It doesn't make sense (according to the above 6/7 also is a terminating decimal because 7 = 2^1 + 5^1, but it isn't).gmatguy16 wrote: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
Calista.
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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.
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.
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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.
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.
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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).
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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
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
simplyjat