MathRevolution wrote:

[GMAT math practice question]

If \(x=343y\), where y is a positive integer, and \(\frac{x}{196}\) is a terminating decimal, what is the smallest possible value of \(y\)?

\(A. 1\)

\(B. 3\)

\(C. 5\)

\(D. 7\)

\(E. 9\)

A terminating decimal will occur only when a fraction, reduced to its lowest terms, contains, in its denominator,

only: powers of 2; powers of 5; or powers of 2 and 5 (powers of 10).

Find the prime factors of 196: \(2^27^2\)

Thus \(\frac{x}{2*2*7*7}\) = a terminating decimal

To make \(\frac{x}{2*2*7*7}\) terminate, x must have, at the least, 7*7 as a factor (in order to "cancel" the sevens in the denominator).

Assess 343y. When in doubt about numbers that do not conform to common divisibility rules, start by dividing by 7 (then 11, then 13, etc.).

x = 343y = 7 * 7 * 7 * y

We need only two of those sevens for x in order to create a terminating decimal. But we want to minimize y -- so we maximize x, and "take" all three 7s for x.

x = 343 * y

x = (7 *7 * 7) * y

x = (7 * 7 * 7) = 343

y = 1

Thus

\(\frac{x}{196}

= \frac{7*7*7}{2*2*7*7} = \frac{7}{4}

= 1.75\)

The smallest possible value for y is 1

Answer A

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