xspongebobx wrote:
Would someone be able to explain statement 2? I am unsure how " (a+1)!(a+1)!, must have only 2's or/and 5's in its prime factorization, which is only possible if (a+1)!=2!=2(a+1)!=2!=2 (all other factorials 3!, 4!, 5!, have primes other than 2 and 5 in them)" was derived
Thanks.
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 \(10=2*5\). Other examples of terminating decimals: fractions \(\frac{1}{8}\), \(\frac{1}{25}\), ...
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 the terminating decimal.
Check below for more:
Math: Number TheoryHow to Identify Terminating Decimals on the GMATTerminating Decimals in Data Sufficiency on the GMATTerminating and Recurring DecimalsEven more:
ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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