Note: The quotient terminates with one decimal digit. Fraction * fraction yields more decimal points. 0.2 * 0.2 = 0.04 and 2 * 0.2 = 0.4. we have 0.4 type result. thus one must be integer and the other is fraction.

Given,

\(\frac{3^x 5^2}{3^55^3}\)

= \(3^{(x-5)} 5^{(2-3)}\)

= \(3^{(x-5)} 5^{(-1)}\)

=\(3^{(x-5)}* \frac{1}{5}\)

\(\frac{1}{5} = 0.2\)

Thus , \(3^{(x-5)}\) has to an integer.

\(x\geq5\) is a must to have integer in this case.

Thus the best answer is D.

\(\frac{3^x5^2}{3^55^3}\) = \(\frac{3^x}{3^55}\). Since \(\frac{1}{5}\) =0.2. Inorder for a fraction to have a terminating decimal, the denominator should only consist of either 2s or 5s.

Hence \(x\geq{5}\).
D is the answer.

Thanks Selim for the explanation.
I also solved the problem but by substituting different values of x (x>0). It took a bit longer time as I couldn't think of the logic of 3 ^(x-5) to be an integer.
How to develop such logics ? Pls suggest.

BTW, my answer- fraction will be reduced to \(3^{(x-5)}* \frac{1}{5}\)
If x=1, \(1/(3)^{4}* 1/5\), since 1/3 itself is non terminating so the whole fraction is non-terminating
If x=2, x=3, x=4, same issue as in x=1
If x=5, fraction =1/5 =0.2; so x can be 5
If x=6, fraction = 3/5 = 3 x 0.2 = 0.6, so x can be 6
If x=7, fraction = 9/5 = 9 x 0.2 = 1.8; so x can be 7
So on so forth, we can conclude from the above results that x>=5. Hence, answer is D

\(\frac{3^x(5^2)}{3^5(5^3)}\), this quotient terminates , implies that x is an integer.

You know 2 or 5 or both 2 and 5, in the denominator always terminates. When a multiple of 3 is divided by 3, then the quotient is never terminated.

So, as long as \(3^5\) is active in the denominator, the quotient won't terminate. We need to cancel out \(3^5\) in the denominator in order to get a terminating decimal of the quotient.

So, x=5 satisfies the condition. however, we need to find out the MUST BE TRUE condition, which is true at all points of the condition rather than at a single point.

So, \(x\geq5\)

Ans. (D)
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Bro , i applied the same concept as u described above. I did it directly. If u put any value for x less than 5 it gives us non terminating decimal but we need one terminating decimal. Thus i said that 3^{(x-5)} has to be an integer. Any value of x less than 5 will give us negative exponent but if \(x \geq5\) we can't avoid the problem.

Thanks.

We can substitute the value of x by using the options..
The value of x less than 5 will not satisfy the given condition
So option d is correct

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