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The integer a is greater than 1 and is not equal to the squ
[#permalink]
21 Jun 2017, 13:48

Expert Reply

Question Stats:

The integer a is greater than 1 and is not equal to the square of an integer. Which of the following answer choices could potentially be equal to the square of an integer?

Indicate all that apply.

❑ \(\sqrt{a}\)

❑ \(a^2\) \(- 1\)

❑ \(a^2\) \(+ 1\)

❑ \(a^2\) \(- a\)

❑ \(a^2\) \(- 2a + 1\)

❑ \(2a\)

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Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
12 Oct 2017, 02:04

Carcass wrote:

The integer a is greater than 1 and is not equal to the square of an integer. Which of the following answer choices could potentially be equal to the square of an integer?

Indicate all that apply.

❑ \(\sqrt{a}\)

❑ \(a^2\) – 1

❑ \(a^2\) + 1

❑ \(a^2\) – a

❑ \(a^2\) – 2a + 1

❑ \(2a\)

Show: :: OA

E,F

What's the best strategy here?

I got the right answers but I decided to stop because I did not find examples for the other choices but it is not the best way to deal with it.

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
20 Feb 2018, 10:08

1

What the question ask we find is x^2 which equals x*x.

A is not a square of a number;thus a is also not a quadruple power of any number;therefore, sqrt a can not be x^2. A is not correct.

Only the square of 0 and square of 1 and -1 is 1 away from each other;therefore B,C are not correct.

a^2-a=a*(a-1) doesn't match witch x*x. D is also not correct.

a^2-2a+1=(a-1)(a-1)=(a-1)^2 E is correct.

a could equal to 2. 2*2=2^2 F is correct

A is not a square of a number;thus a is also not a quadruple power of any number;therefore, sqrt a can not be x^2. A is not correct.

Only the square of 0 and square of 1 and -1 is 1 away from each other;therefore B,C are not correct.

a^2-a=a*(a-1) doesn't match witch x*x. D is also not correct.

a^2-2a+1=(a-1)(a-1)=(a-1)^2 E is correct.

a could equal to 2. 2*2=2^2 F is correct

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
23 Feb 2018, 22:53

I don't understand why E is the correct answer. Maybe I don't clearly understand the question, but I used the plugging in method to solve this problem.

I have a=2 and found that only 2a (F) works.

I have a=2 and found that only 2a (F) works.

hugebigmac wrote:

What the question ask we find is x^2 which equals x*x.

A is not a square of a number;thus a is also not a quadruple power of any number;therefore, sqrt a can not be x^2. A is not correct.

Only the square of 0 and square of 1 and -1 is 1 away from each other;therefore B,C are not correct.

a^2-a=a*(a-1) doesn't match witch x*x. D is also not correct.

a^2-2a+1=(a-1)(a-1)=(a-1)^2 E is correct.

a could equal to 2. 2*2=2^2 F is correct

A is not a square of a number;thus a is also not a quadruple power of any number;therefore, sqrt a can not be x^2. A is not correct.

Only the square of 0 and square of 1 and -1 is 1 away from each other;therefore B,C are not correct.

a^2-a=a*(a-1) doesn't match witch x*x. D is also not correct.

a^2-2a+1=(a-1)(a-1)=(a-1)^2 E is correct.

a could equal to 2. 2*2=2^2 F is correct

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
24 Feb 2018, 02:18

1

Expert Reply

You cannot pick two because the stem says: is not equal to the square of an integer.

Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well.

6 and 8, for instance, are good numbers to pick.

Hope this helps

_________________

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Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well.

6 and 8, for instance, are good numbers to pick.

Hope this helps

_________________

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Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
24 Feb 2018, 02:36

I see now what you mean. This type of question can be tricky because of the wording. Thanks

Carcass wrote:

You cannot pick two because the stem says: is not equal to the square of an integer.

Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well.

6 and 8, for instance, are good numbers to pick.

Hope this helps

Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well.

6 and 8, for instance, are good numbers to pick.

Hope this helps

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
24 Feb 2018, 09:58

1

gremather wrote:

I don't understand why E is the correct answer. Maybe I don't clearly understand the question, but I used the plugging in method to solve this problem.

I have a=2 and found that only 2a (F) works.

I have a=2 and found that only 2a (F) works.

hugebigmac wrote:

What the question ask we find is x^2 which equals x*x.

A is not a square of a number;thus a is also not a quadruple power of any number;therefore, sqrt a can not be x^2. A is not correct.

Only the square of 0 and square of 1 and -1 is 1 away from each other;therefore B,C are not correct.

a^2-a=a*(a-1) doesn't match witch x*x. D is also not correct.

a^2-2a+1=(a-1)(a-1)=(a-1)^2 E is correct.

a could equal to 2. 2*2=2^2 F is correct

A is not a square of a number;thus a is also not a quadruple power of any number;therefore, sqrt a can not be x^2. A is not correct.

Only the square of 0 and square of 1 and -1 is 1 away from each other;therefore B,C are not correct.

a^2-a=a*(a-1) doesn't match witch x*x. D is also not correct.

a^2-2a+1=(a-1)(a-1)=(a-1)^2 E is correct.

a could equal to 2. 2*2=2^2 F is correct

2*2-2*2+1=1 which equals 1^2 so e is correct.

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
10 Apr 2018, 03:21

3

E is the identity a^2 - 2*a*b + b^2 = (a - b)^2 where a is a and b is 1.

for b and c options, we can just see that no two squares are consecutive numbers, hence we can rule em out

for b and c options, we can just see that no two squares are consecutive numbers, hence we can rule em out

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
11 Apr 2018, 19:35

Is plugging some integers only option to solve this? If yes then how do we decide which number to select.

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
12 Apr 2018, 05:58

Expert Reply

If you follow the stem, then it is very clear.

You have to pick a number > 1 and that is not a square of an integer.

I.E 2 and 4 NO because they are square: 4 is the square of 2.

9 neither, is the square of 3. As such, 3 and 9 are out..

And so forth

_________________

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You have to pick a number > 1 and that is not a square of an integer.

I.E 2 and 4 NO because they are square: 4 is the square of 2.

9 neither, is the square of 3. As such, 3 and 9 are out..

And so forth

_________________

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Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
24 May 2018, 10:26

Carcass wrote:

If you follow the stem, then it is very clear.

You have to pick a number > 1 and that is not a square of an integer.

I.E 2 and 4 NO because they are square: 4 is the square of 2.

9 neither, is the square of 3. As such, 3 and 9 are out..

And so forth

You have to pick a number > 1 and that is not a square of an integer.

I.E 2 and 4 NO because they are square: 4 is the square of 2.

9 neither, is the square of 3. As such, 3 and 9 are out..

And so forth

2 is not square of any integer. Then why can't we take 2?? Please clarify my doubt.

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
26 May 2018, 01:14

Expert Reply

You cannot pick two because the stem says: is not equal to the square of an integer.

Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well.

6 and 8, for instance, are good numbers to pick.

_________________

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Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well.

6 and 8, for instance, are good numbers to pick.

_________________

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Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
11 Dec 2018, 17:37

I misunderstood the potentially part so I totally skipped F. IS there any other logical solutions besides plugging in?

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
12 Dec 2018, 04:20

1

Expert Reply

I think no in this specific case.

However, that is me.

Regards

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However, that is me.

Regards

_________________

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Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
30 Apr 2019, 02:59

Carcass wrote:

You cannot pick two because the stem says: is not equal to the square of an integer.

Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well.

6 and 8, for instance, are good numbers to pick.

Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well.

6 and 8, for instance, are good numbers to pick.

I am really sorry, but I still cant wrap my head around why 2 cannot be used. If we follow your advice then even 6 has a square 36. So I wouldn't pick both 36 and 6, in the same way I wouldn't pick 4 and 2.

OP says "is not equal to the square of an integer" - 2 is the square of ~1.414, which is a non integer.

"Which means 4 is the square of two so you cannot pick four. But is also true the other way around: cannot pick two as well." - this is true for any integer, because any integer will have some square (which is itself an integer). I am confused about the other way round part.

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
01 May 2019, 11:36

Expert Reply

On the number line if you do list the numbers > 1

2,3,4,5,6,7,8,9,10,11,12,13,14,..............

4 is the natural square of 2

However, 5 is NOT a square of any number

6 as well

7 as well

8 as well

9 yes, is the square of 3

Hope this helps

_________________

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2,3,4,5,6,7,8,9,10,11,12,13,14,..............

4 is the natural square of 2

However, 5 is NOT a square of any number

6 as well

7 as well

8 as well

9 yes, is the square of 3

Hope this helps

_________________

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Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
13 Jun 2019, 12:19

If we pick 5 then-

2a = 10 ---> which wont be the square of an integer. But the OA says E and F both.

2a = 10 ---> which wont be the square of an integer. But the OA says E and F both.

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
19 Jul 2020, 11:03

1

If you can't use 2 (which I don't understand, I think you can use 2 as the stem says N>1 and not a square of a number, which would not rule out 2), you have to go all the way up to 18 to find 2a = 36, which is 6*6, therefore F is also correct

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
24 Sep 2020, 17:31

Carcass wrote:

The integer a is greater than 1 and is not equal to the square of an integer. Which of the following answer choices could potentially be equal to the square of an integer?

Indicate all that apply.

❑ \(\sqrt{a}\)

❑ \(a^2\) \(- 1\)

❑ \(a^2\) \(+ 1\)

❑ \(a^2\) \(- a\)

❑ \(a^2\) \(- 2a + 1\)

❑ \(2a\)

Awesome question.

First we need to establish the numbers we can plug in for \(a\).

If a number \(x\) is not a square of an integer, then \(x ≠ y^2\), where \(y\) is any integer. So plug in integers for \(y\), and they can't be \(a\).

4,9,16,25,36..... can't be used.

The question is, if we plug in any other integers for \(a\), will they result in a square of an integer? (4,9,16,25,36....)

❑ \(\sqrt{a}\)

This is incorrect, since the only way this yields an integer is if \(a\) is a square of an integer, and we know it can't be.

____________

❑ \(a^2\) \(- 1\)

This is incorrect because no square of an integer exists that is one integer away on the number line from it, other than 1.

Think \(2^2 - 1 = 3\), or \(3^2 - 1 = 8\). We have our square already, and we're subtracting one away from it, so the result can't be a square.

If \(a = 1\), then it could be, since:

\(1^2 - 1 = 0\)

And 0 is a square of an integer ( \(\sqrt{0} = 0\) ). But we know that \(a ≠ 1\), so this choice must be incorrect.

____________

❑ \(a^2\) \(+ 1\)

This is incorrect for the same reason that the above is incorrect. This time, not even \(a = 1\) helps this one.

____________

❑ \(a^2\) \(- a\)

Again, using similar logic as above, this is incorrect. However 1 would work for this case:

\(1^2 - 1 = 0\)

____________

❑ \(a^2\) \(- 2a + 1\)

This one is correct. To see it, you have to factor it:

\((a-1)^2\)

Now we can let \(a = 5\), and we get 16, and 16 is a square of an integer.

____________

❑ \(2a\)

This one is also correct. Plug in 8 to get:

\(2(8) = 16\), and 16 is a square of an integer.

____________

So the answers are E and F

gmatclubot

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