Terminating Decimals Problem - Manhattan

[user4544]

I don’t understand the solution in Manhattan book, p.452 #32.

The question is: If 3^x(5^2) is divided by 3^5(5^3), the quotient terminates with one decimal digit. If x>0, which of the following statements must be true

a) x is even
b) x is odd
c) x<5
d) x greater than or equal to 5
e) x =5

I completely understand that to have a terminating decimal, you need to have a 2 or 5 in the denominator and no other number otherwise, the decimal will be repeating. Thus, in this case we have to eliminate the 3 on the denominator so that there is only 5s on the denominator and the number terminates.

My understanding, was that if x=5, then it will cancel the 3^5 on the denominator and we have a one decimal digit number that terminates. I see that the answer is D, but my initial hesitation with D was the “one decimal digit” part of the question. How am I supposed to know that 3 raised to the power of any number above 5, will still terminate in one decimal digit?

[cylverixxx]

@user4544

How am I supposed to know that 3 raised to the power of any number above 5, will still terminate in one decimal digit?

\frac n5 (where n is an integer) can be rewritten as \frac{2n}{10}, and you know this results in a decimal with at most one decimal place regardless of what n is.

Owing to what was mentioned above, \frac{3^x}{5} is bound to terminate with one decimal place (can’t terminate with 0 dp because 5 isn’t a divisor of 3^x) for any natural number x.

@user4544

My understanding, was that if x=5, then it will cancel the 3^5 on the denominator and we have a one decimal digit number that terminates.

You could’ve just tried x = 6 and easily surmised that 3/5 = 0.6 also works (terminates with one decimal place), thus disproving E.