I. \((\sqrt{82} + \sqrt{82})^2\)
= \((2\sqrt{82})^2\)
= 4 * 82 = 328 integer

2. \(82\sqrt{82}\)
cannot be expressed as integer

3. \(\frac{\sqrt{82}\sqrt{82}}{82}\)
= 82/82
= 1, which is integer

Answer option E

1. ( (82)^(1/2) + (82)^(1/2) )^2
= ( 2 *(82)^(1/2))^2
= 4 * 82
= 328
Integer

2. 82 (82)^(1/2)
can't be written as an integer as 82 is not a perfect square .

3. ((82)^(1/2) * (82)^(1/2) )/82
= 1
Integer

Answer E
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1st:

[SQRT(82) + SQRT(82)]^2

we have formula (a+b)^2 = a^2 +b^2 +2ab
we can rewrite the first statement as 82+82 + 2*[SQRT(82)*SQRT(82)]
[SQRT(82)*SQRT(82)] this is equal to 82. we get an integer.

We can eliminate
(A) None
(C) III only

Statement II:
82*sqrt(82)
we know that 82 is not a perfect square, thus, it is not an integer. Statement II will not yield an integer.
We can eliminate:

(D) I and II


Statement III:
[sqrt(82) * sqrt (82)]/82
sqrt(82) multiplied by itself will result in 82. 82 divide by 82, and get 1. This is an integer.

Eliminate answer choice B, and the only answer choice left, the correct one is E.

l can be written as (2√82)^2 ..a integer
lll will equal to 1 so Option E is correct answer.

I. We use the identity (a+b)^2 = a^2+2ab+b^2 which gives us:
82 + (2*\(\sqrt{82}\)*\(\sqrt{82}\)) + 82
Since \(\sqrt{82}\)*\(\sqrt{82}\) equals 82 this will yield an integer.

II. No since \(\sqrt{82}\) is not a perfect square and you cannot separate it either into to two perfect squares.


III. Since \(\sqrt{82}\)*\(\sqrt{82}\) equals 82 this will yield an integer : 82/82=1


Answer E : I and III only

Attached is a visual that should help.

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lets see the options =>
Option 1=> (2√82)^2=> 4*82 => integer
Option 2 => 82√82 => non integer
Option 3 => √82*√82 / 82 => 82/82=> 1 , which is off course an integer.
Hence E
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We'll use the fact that (√82)(√82) = 82

I. (√82 + √82)² = (2√82)² = (2√82)(2√82) = (2)(2)(√82)(√82) = (4)(82) = some integer
II. Doesn't look like it could be an integer. I'm not sure, so I'll move onto to statement III
III. (√82)(√82)/82 = 82/82 = 1

NOTE: As we check each statement, we should also check the answer choices.
Since I and III both work, the correct answer must be E, since there's no option for all 3 to be true.

Answer: E
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i. \({ \left( \sqrt { 82 } +\sqrt { 82 } \right) }^{ 2 }=\left( 2\sqrt { 82 } \right) ^{ 2 }\Rightarrow\) At this moment you already know you have an integer without further calculations.

ii. \(82\sqrt { 82 } \Rightarrow\) We also know that perfect squares don't end in 2, 3, 7 or 8. So, rule this one out as this product won't yield an integer.

iii. \(\cfrac { \sqrt { 82 } \sqrt { 82 } }{ 82 } =\quad \cfrac { \sqrt { 82\ast 82 } }{ 82 } =\cfrac { \sqrt { { 2 }^{ 2 }\ast { 41 }^{ 2 } } }{ 2\ast { 41 } } =\cfrac { 2\ast 41 }{ 2\ast 41 } =\quad 1\quad \Rightarrow \quad integer\)

OA
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Let’s simplify each Roman numeral:

I. (√82 + √82)^2

(√82 + √82)^2 = (2√82)^2 = 4 x 82 = 328

We see that this an integer.

II. 82√82

Since √82 is a non-terminating decimal, 82√82 is not an integer.

III. (√82√82)/82

(√82√82)/82 = 82/82 = 1

We see that this is an integer.

Answer: E
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Which of the following expressions can be written as an integer?

I. \((\sqrt{82} + \sqrt{82})^2\)
82 + 82 + 2*82 = 4*82 = 328
INTEGER

II. \(82\sqrt{82}\)
82 * \sqrt{82}
9<\sqrt{82}<10 and is NOT an integer
NOT AN INTEGER

III. \(\frac{\sqrt{82}\sqrt{82}}{82}\)
82/82 = 1
INTEGER

IMO E
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Solution:

We can easily and immediately observe III to be = 82/82 =1 (Eliminate A,B,D)

I - (√82+√82)^2 = (2√82)^2

= 4 X 82 =An integer (eliminate C)

(option e)
Devmitra Sen
GMAT SME
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Answer: Option E

Video solution by GMATinsight



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