If d is a positive integer, is d^1/2 an integer?

SOLUTION

If d is a positive integer, is \(\sqrt{d}\) an integer?

Given: \(d=integer\).
Question: is \(\sqrt{d}=integer\)? or: is \(d\) a perfect square: is \(d=integer^2\)?

(1) \(d\) is the square of an integer --> \(d=integer^2\) --> directly gives an answer. Sufficient.

(2) \(\sqrt{d}\) is the square of an integer --> \(\sqrt{d}=integer^2=integer\) (as square of an integer is an integer). Sufficient.

Answer: D.

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\(\)If d is a positive integer, is \(\sqrt{d}\) an integer?

(1) d is the square of an integer.
(2) \(\sqrt{d}\) is the square of an integer.

Sol: Given d is a postive integer we need to check whether \(\sqrt{d}\) = I where I is an integer

st1: Given d=a^2 where " a" is some postive integer then \(\sqrt{d}\) =a which is an integer So St 1 is sufficent

St2: given \(\sqrt{d}\) = b^2 where b is some integer so \(\sqrt{d}\) is an integer.

Ans is D.

Note: • When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \sqrt{[4]{x}}, then the only accepted answer is the positive root.

That is, \(\sqrt{25}\) , NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

• Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).


Courtesy GMAT CLUB Math Book
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A.) d = x^2
Therefore, root[d] = x
e.g. d= 16 then root[d] = 4

SUFFICIENT

B.) root[d] = x^2
square of an integer is an integer
e.g. 2^2 = 4, 3^2 = 9
e.g root[d] = 2^2 = 4

SUFFICIENT

Answer - D

Difficulty level - 550

Here is another way of attacking this question. This different approach applies mostly to statement 2.

d is a positive integer, is (d^1/2) an integer?
rephrased, is d a perfect square?

1) d=x^2
square root of x^2 will be d (a positive integer) SUFFICIENT

2) d^(1/2)=x^2
square both sides
d=x^4
any number raised to an even root is a perfect square. square root of x^4 will always be an integer. SUFFICIENT

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