If x=343y, where y is a positive integer,

Max@Math Revolution wrote:[GMAT math practice question]

$$If\ \ x=343y,\ where\ y\ is\ a\ positive\ integer,\ and\ \frac{x}{196}\ is\ a\ ter\min ating\ decimal,\ what\ is\ the\ smallest\ possible\ value\ of\ y?$$

A. 1
B. 3
C. 5
D. 7
E. 9

We may recall that a fraction will be a terminating decimal when the denominator of its lowest terms has only prime factors of 2 and/or 5. .

We see that 196 = 98 x 2 = 49 x 2 x 2 = 7^2 x 2^2

In order for x/196 to be a terminating decimal, the numerator x must contain at least the number 49 = 7^2, thereby canceling out the 7^2 in the denominator..
We see that 343 = 7^3, so if y = 1, x = 343 and thus 343/196 will be a terminating decimal.

Answer: A

=>

$$\frac{x}{196}=\frac{\left(343\cdot y\right)}{196}=\frac{\left(7^3\cdot y\right)}{14^2}=\frac{\left(7^3\cdot y\right)}{2^2\cdot7^2}=\frac{7y}{4}$$
As the denominator has only 2 as a prime factor, it is a terminating decimal, regardless of the value of y.
Thus, the smallest possible value of y is 1.

Therefore, the answer is A.

Answer: A