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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$6$

Vedclass · Answer

(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.

Similar Questions

Write whether every positive integer can be of the form $4q + 2$,where $q$ is an integer. Justify your answer.

If $k_{1}$ and $k_{2}$ are two distinct prime integers,then their $\text{l.c.m.}$ is ..........

$LCM(15, 21, 35) = .........$

The smallest positive number divisible by every integer from $2$ to $10$ is:

Find $g.c.d.$ of $210$ and $55$ by Euclid's algorithm.

Vedclass Test Series

Exam Paper Generator

Online Exam Module