Question

the decimal representation of 14587/2^1×5^4 will terminate after how many decimal place..??​

Answer 1

Answer:

We are given:

$\to\sf \dfrac{14587}{2^1 \times 5^4}$

Multiply 2³ in numerator and denominator.

$\to\sf \dfrac{14587}{2^1 \times 5^4} \times \dfrac{2^3}{2^3}$

$\to\sf \dfrac{14587 \times 2^3}{2^1 \times 5^4 \times 2^3}$

$\to\sf \dfrac{14587 \times 2^3}{ \left(2^1 \times 2^3 \right) \times 5^4 }$

$\to\sf \dfrac{14587 \times 2^3}{ 2^4 \times 5^4 }$

$\boxed{\sf{\red{Use \ a^m \times b^m = (ab)^m}}}$

$\to\sf \dfrac{14587 \times 2^3}{ 10^4 }$

Now, denominator is 10⁴. The exponent "4" will tell the value in which the number ends.

Thus, the answer is 4.

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Answer 2

SOLUTION

TO DETERMINE

The decimal representation of the below rational number will terminate after how many places of decimal

$\displaystyle \sf{ \frac{14587}{ {2}^{1} \times {5}^{4} } }$

CONCEPT TO BE IMPLEMENTED

$\displaystyle\sf{Fraction = \frac{Numerator}{Denominator} }$

A fraction is said to be terminating if prime factorisation of the denominator contains only prime factors 2 and 5

If the denominator is of the form

$\sf{Denominator = {2}^{m} \times {5}^{n} }$

Then the fraction terminates after N decimal places

Where N = max { m , n }

EVALUATION

Here the given rational number is

$\displaystyle \sf{ \frac{14587}{ {2}^{1} \times {5}^{4} } }$

Numerator = 14587

$\displaystyle \sf{ Denominator = {2}^{1} \times {5}^{4} }$

Since the prime factorisation of the denominator contains only prime factors as 2 and 5

So the given rational number is terminating

The exponent of 2 = 1

The exponent of 5 = 4

Noe max { 1 , 4 } = 4

Hence the given rational number terminates after 4 decimal places

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