Confused - Any Ideas

[phoenix9801]

Problem 6:

n= Square root (2/7+1/3+4/9+2/3+5/9+5/7+x)

In the equation above, if n is an integer, which of the following could be a possible value of x?

Indicate all such values:

A) 0
B) 2/63
C) 1
D) 65/63
E) 4

the way I solved it was reorder the fractions as follows

n = Square root ((1/3+2/3)+(3/7+5/7)+(4/9+5/9)+x) =
n = Square root (1+1+1+x)
n = Square root (3+x)

But the problem for me is i do not know what to do next? to answer the chooses.
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[jrakhe]

As per your solution if X=1 then n will be 2 which is an integer.
So answer is X=1

[phoenix9801]

Can you explain in detail please how to get from n= Square root(3+x) to x = 1 ????

As per your solution if X=1 then n will be 2 which is an integer.
So answer is X=1

[phoenix9801]

Can Anyone help me for further explanation in greater detail please.

[gmacpup]

phoenix,
your working is correct, got to test it with the available options.
Say, Sqr root of (3+X)

first, substitute 0 for X which would be SR of 3 - not an integer.
second value, x=2/65 results in 3 + 0.0(value) - again not an integer
if x=1, then SR of 3+1=4 is 2 - an integer.
Hence the answer is 1.

[Stuart@KaplanGMAT]

Can you explain in detail please how to get from n= Square root(3+x) to x = 1 ????

From the question stem, we know that n is an integer. Accordingly, we know that sqrt(3+x) is an integer. What numbers have roots that are integers? Perfect squares.

Consequently, (3+x) must be a perfect square, i.e. 1, 4, 9, 16, 25, 36, ...

Now we just need to plug in each answer to see which ones give us perfect squares (on the GMAT you'll never see "indicate all such values" questions - is this a GRE question? Please always indicate the source!).

A) 3+0 = 3... not a perfect square.
B) 3+ (2/63) = fraction... not a perfect square.
C) 3 + 1 = 4... ding ding ding!
D) 3 + 65/63 = fraction... not a perfect square.
E) 3 + 4 = 7... not a perfect square.

Only (C) gives us a perfect square: choose C!