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If x=343y, where y is a positive integer, and x/196 is a

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If x=343y, where y is a positive integer, and x/196 is a [#permalink]

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New post 23 Jan 2018, 00:11
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[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 \(y\)?

\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)
[Reveal] Spoiler: OA

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If x=343y, where y is a positive integer, and x/196 is a [#permalink]

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New post 23 Jan 2018, 00:27
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 \(y\)?

\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)


Easy question if we think this way -
Given, \(x = 7^3y.\)... or\(\frac{x}{49} = 7y\).... ---> as \(\frac{x}{49*4} = \frac{7y}{4}\). Now, the samllest possible value will be \(y = 1\).
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If x=343y, where y is a positive integer, and x/196 is a [#permalink]

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New post 23 Jan 2018, 16:53
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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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Re: If x=343y, where y is a positive integer, and x/196 is a [#permalink]

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New post 24 Jan 2018, 09:34
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 \(y\)?

\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)


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
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Re: If x=343y, where y is a positive integer, and x/196 is a [#permalink]

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New post 25 Jan 2018, 03:06
=>

\(\frac{x}{196} = \frac{(343*y)}{196} = \frac{(7^3*y)}{(14^2)} = \frac{(7^3*y)}{(2^2)(7^2)} = \frac{(7y)}{4}\)

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
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Re: If x=343y, where y is a positive integer, and x/196 is a [#permalink]

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New post 25 Jan 2018, 10:23
7y/4 7/4 is a terminating decimal. so y=1
Re: If x=343y, where y is a positive integer, and x/196 is a   [#permalink] 25 Jan 2018, 10:23
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