The decimal expansion of frac{9}{1600} will t | vedclass

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(A) To find the number of decimal places after which the expansion of $\frac{9}{1600}$ terminates,we first express the denominator in prime factors.$1

Vedclass · Accepted Answer

$6$

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(A) To find the number of decimal places after which the expansion of $\frac{9}{1600}$ terminates,we first express the denominator in prime factors.$1600 = 16 	imes 100 = 2^4 	imes 10^2 = 2^4 	imes (2 	imes 5)^2 = 2^4 	imes 2^2 	imes 5^2 = 2^6 	imes 5^2$.Now,the fraction is $\frac{9}{2^6 	imes 5^2}$.To make the powers of $2$ and $5$ equal,we multiply the numerator and denominator by $5^4$:$\frac{9 	imes 5^4}{2^6 	imes 5^2 	imes 5^4} = \frac{9 	imes 625}{2^6 	imes 5^6} = \frac{5625}{(2 	imes 5)^6} = \frac{5625}{10^6}$.Since the denominator is $10^6$,the decimal expansion will terminate after $6$ digits.

The decimal expansion of $\frac{9}{1600}$ will terminate after $\ldots \ldots \ldots \ldots$ digits.

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