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22 May 2010, 14:35
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A terminating decimal is defined as a decimal that has a finite number of nonzero digits. Examples of terminating decimals are 0.24, 52, and 6.0314. x and y are positive integers. If x/y is expressed as a decimal, is it a terminating decimal?
(1) 40 < x < 45
(2) y = 8
(1) 40 < x < 45
(2) y = 8
22 May 2010, 16:46
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Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.
(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)
In original question statement (2) says that denominator equals to 2^3=8, hence x/8 will be terminating decimal no matter what the value of x is.
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.
(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)
In original question statement (2) says that denominator equals to 2^3=8, hence x/8 will be terminating decimal no matter what the value of x is.
24 Apr 2012, 04:20
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Found this helpful explanation on Mgmat site. 
Author- Emily Sledge
This rule took a while for me to internalize. It's tough to picture a decimal terminating when the denominator is so huge, such as DWG's example of 43/256. I found it helped me to think about the basic patterns:
1/2^1 = 0.5
1/2^2 = 0.25
1/2^3 = 0.125
1/2^4 = 0.0625
1/2^5 = 0.03125
1/2^6 = 0.015625
1/2^7 = 0.0078125
1/5^1 = 0.2
1/5^2 = 0.04
1/5^3 = 0.008
1/5^4 = 0.0016
1/5^5 = 0.00032
1/5^6 = 0.000064
1/5^7 = 0.0000128
Every one of these terminates, and the pattern indicates that would continue to be true for higher powers. The number of decimal places increases along with the powers of 2 or 5, but the number of decimal places will always be finite.
In contrast, any factors other than 2 or 5 in the denominator can quickly be shown to be non-terminating, even for the most basic case (exponent of 1). Higher powers would be even messier:
1/3 = 0.33333(3 repeating)
1/6 = 0.16666(6 repeating)
1/7 = 0.142857(142857 repeating)
1/9 = 0.11111(1 repeating)
Author- Emily Sledge
This rule took a while for me to internalize. It's tough to picture a decimal terminating when the denominator is so huge, such as DWG's example of 43/256. I found it helped me to think about the basic patterns:
1/2^1 = 0.5
1/2^2 = 0.25
1/2^3 = 0.125
1/2^4 = 0.0625
1/2^5 = 0.03125
1/2^6 = 0.015625
1/2^7 = 0.0078125
1/5^1 = 0.2
1/5^2 = 0.04
1/5^3 = 0.008
1/5^4 = 0.0016
1/5^5 = 0.00032
1/5^6 = 0.000064
1/5^7 = 0.0000128
Every one of these terminates, and the pattern indicates that would continue to be true for higher powers. The number of decimal places increases along with the powers of 2 or 5, but the number of decimal places will always be finite.
In contrast, any factors other than 2 or 5 in the denominator can quickly be shown to be non-terminating, even for the most basic case (exponent of 1). Higher powers would be even messier:
1/3 = 0.33333(3 repeating)
1/6 = 0.16666(6 repeating)
1/7 = 0.142857(142857 repeating)
1/9 = 0.11111(1 repeating)
