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

Hello Bunuel

Your explanation is to the point but in the official explanation, its mentioned:

82* 82^3/2

82^3= 2^3 *41^3

How did we get this cube?

Could you please post a screenshot or post entire solution?

Expression II does not represent an integer because (82)* 82^1/2= 82^3/2 and 82^3= 2^3* 41^3 is not a perfect square. Regarding this last assertion, note that the square of any integer has the property that each of its distinct prime factors is repeated an even number of times.



(Explanation taken from official guide 2017)

\(82*82^{(\frac{1}{2})}=82^{(1+\frac{1}{2})}=82^{\frac{3}{2}}=\sqrt{82^3}\)

Let’s simplify each expression:

1. (√82 + √82)^2

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

We see that this an integer.

2. 82√82

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

3. (√82√82)/82

(√82√82)/82 = 82/82 = 1

We see that this is an integer.

Answer: E

soving the expression only 1 & 3 gives integer values.

1. use (a+b)^2. = integer
2. sq. root a * a= a*sqroot a= not integer
3 sq root a * sq. root a = a

Answer E

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
III. (√82)(√82)/82 = 82/82 = 1

NOTE: As we're checking expressions, we should also be checking 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.

Cheers,
Brent

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

1. \((\sqrt{82}+\sqrt{82})^2\)

2. \(82∗\sqrt{82}\)

3. \(\frac{\sqrt{82}∗\sqrt{82}}{82}\)

1) \((\sqrt{82}+\sqrt{82})^2\)= \((2*\sqrt{82})^2\)=4*82= Integer

2) \(82∗\sqrt{82}\)= 82 * Non integer= Non-integer

3) \(\frac{\sqrt{82}∗\sqrt{82}}{82}\)=\(\frac{(\sqrt{82}∗\sqrt{82})}{(\sqrt{82}∗\sqrt{82})}\)= 1= integer



A. none
B. 1 only
C. 3 only
D. 1 and 2
E. 1 and 3 Answer

Hi All,

We’re asked which of the following Roman Numerals can be written as an INTEGER.

While Roman Numeral questions can sometimes be time-consuming, this one is relatively straight-forward and is built on a couple of Number Property rules that can help you to avoid doing heavy calculations. You also do not need to consider all 3 Roman Numerals to answer it…

First, it’s worth noting that 82 is NOT a Perfect Square (if you know your Perfect Squares at this point, then you know that 9x9 = 81 and 10x10 = 100 are the two Squares just ‘below’ and ‘above’ 82, respectively).

I.

Adding any two identical terms together can be rewritten as 2(that term). For example, 3 + 3 = 2(3)…. X + X = 2(X)…. Etc.

Here, we’re adding √82 to √82, so that’s 2(√82). Squaring THAT term would give us (2)(2)(82)… which is clearly an INTEGER (it’s 328, but you don’t actually have to do that math).

III.

Here, we are multiplying the same Square Root against itself – and any time you do that, you end up with the original integer. For example (√4)(√4) = (2)(2) = 4… (√5)(√5) = 5… Etc.

Thus, we would end up with 82 in the numerator and 82 in the denominator – and 82/82 = 1. This also is clearly an INTEGER.

Based on how the Answers are written, we can stop working.

Final Answer:

GMAT Assassins aren’t born, they’re made,
Rich

Option I: \([\sqrt{82} + \sqrt{82}]^2 = [2\sqrt{82}]^2 = 4 * 82\) - Integer

OptionII: \(82 ∗ \sqrt{82}\) - since \(\sqrt{82}\) is not an integer then option II is not integer either.

Option III: \(\sqrt{82} * \sqrt{82}\) = \(82\) divided by \(82 = 1\) - Integer.

Only Option I and III.

ANswer E

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