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M04-26

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M04-26  [#permalink]

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New post 16 Sep 2014, 00:23
4
18
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

53% (01:14) correct 47% (01:09) wrong based on 169 sessions

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Re M04-26  [#permalink]

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New post 16 Sep 2014, 00:23
2
1
Official Solution:


Is \(\frac{7^7}{7^x}\) an integer?

\(\frac{7^7}{7^x} = 7^{7-x}\), so as long as \(x\) is an integer and \(x \le 7\), the expression is an integer.

(1) \(0 \le x \le 7\). If \(x\) is an integer from this range, then \(\frac{7^7}{7^x}\) will be an integer but if \(x\) is a fraction from this range, then \(\frac{7^7}{7^x}\) will NOT be an integer.

(2) \(|x| = x^2\)

Square \(|x| = x^2\);

\(x^2 = x^4\);

\(x^2(x^2 - 1) = 0\)

\(x^2(x - 1)(x + 1) = 0\);

\(x = 0\), 1, or -1. For each of those values \(\frac{7^7}{7^x}\) IS an integer. So, we have an YES answer.


Answer: B
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Re: M04-26  [#permalink]

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New post 26 Nov 2014, 13:20
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Bunuel wrote:
Official Solution:


\(\frac{7^7}{7^x} = 7^{7-x}\), so as long as \(x\) is an integer and \(x \le 7\), the expression is an integer.

Statement (1) by itself is insufficient. S1 says that \(x\) can be between 0 and 7, so it can be an integer or any fraction.

Statement (1) by itself is sufficient. S2 implies that \(x\) is one of (-1, 0, 1).


Answer: B


Hi Bunnel,
S2 is found by substituting values or by any other way ? Any other way available other than substituting ??

Thanks,
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Re: M04-26  [#permalink]

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New post 27 Nov 2014, 06:49
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prashd wrote:
Bunuel wrote:
Official Solution:


\(\frac{7^7}{7^x} = 7^{7-x}\), so as long as \(x\) is an integer and \(x \le 7\), the expression is an integer.

Statement (1) by itself is insufficient. S1 says that \(x\) can be between 0 and 7, so it can be an integer or any fraction.

Statement (1) by itself is sufficient. S2 implies that \(x\) is one of (-1, 0, 1).


Answer: B


Hi Bunnel,
S2 is found by substituting values or by any other way ? Any other way available other than substituting ??

Thanks,


You can solve it algebraically:

Square |x| = x^2;

x^2 = x^4;

x^2(x^2 - 1) = 0

x^2(x - 1)(x + 1) = 0;

x = 0, 1, or -1.

Hope it's clear.
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Re: M04-26  [#permalink]

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New post 28 Nov 2014, 07:31
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need to be careful..........
x is not integer, x can be 1.2, 3.5 5.6 anything. Any decimal in (0,7).....
CAREFUL!!!!!!!!!!!!!!!!!!!!!!!!!!
hence using 1st option it can be fraction!!!! Grrr.....
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Re: M04-26  [#permalink]

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New post 24 Jul 2016, 07:47
Just a small doubt
st 2) three values of x are derived... -1 , 0 ,1
so three different values... and stiff st sufficient?
thanks
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Re: M04-26  [#permalink]

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New post 24 Jul 2016, 08:24
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Re: M04-26  [#permalink]

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New post 07 Oct 2016, 17:21
Bunuel wrote:
prashd wrote:
Bunuel wrote:
Official Solution:


\(\frac{7^7}{7^x} = 7^{7-x}\), so as long as \(x\) is an integer and \(x \le 7\), the expression is an integer.

Statement (1) by itself is insufficient. S1 says that \(x\) can be between 0 and 7, so it can be an integer or any fraction.

Statement (1) by itself is sufficient. S2 implies that \(x\) is one of (-1, 0, 1).


Answer: B


Hi Bunnel,
S2 is found by substituting values or by any other way ? Any other way available other than substituting ??

Thanks,


You can solve it algebraically:

Square |x| = x^2;

x^2 = x^4;

x^2(x^2 - 1) = 0

x^2(x - 1)(x + 1) = 0;

x = 0, 1, or -1.

Hope it's clear.



im sorry how was this even calculated?

I think aswer should be E

This is how I see S2:
x is equivalent to its absolute value which is given as x^2

7^7/7^x2


and thats all we get using S2 alone. From s1, we then plug in the values of x. when we reach value 7, 7 square is 49 hence 7^7 / 7^49 is NOT an integer

now obviously i know im wrong since this is not the correct answer but i wanted to show my understanding of the question. can anyone please clarify? thanks
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Re: M04-26  [#permalink]

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New post 08 Oct 2016, 03:07
jonmarrow wrote:
im sorry how was this even calculated?

I think aswer should be E

This is how I see S2:
x is equivalent to its absolute value which is given as x^2

7^7/7^x2


and thats all we get using S2 alone. From s1, we then plug in the values of x. when we reach value 7, 7 square is 49 hence 7^7 / 7^49 is NOT an integer

now obviously i know im wrong since this is not the correct answer but i wanted to show my understanding of the question. can anyone please clarify? thanks


As shown there only 3 values of x satisfy |x| = x^2: 0, 1 and -1. The question asks whether 7^7/7^x is an integer. For each of those values 7^7/7^x IS an integer. So, we have an YES answer.
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Re: M04-26  [#permalink]

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New post 25 Oct 2018, 23:27
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please elaborate on how " S2 implies that xx is one of (-1, 0, 1)."?
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New post 26 Oct 2018, 00:00
rakshithv22 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please elaborate on how " S2 implies that xx is one of (-1, 0, 1)."?


(2) \(|x| = x^2\)

Square \(|x| = x^2\);

\(x^2 = x^4\);

\(x^2(x^2 - 1) = 0\)

\(x^2(x - 1)(x + 1) = 0\);

\(x = 0\), 1, or -1. For each of those values \(\frac{7^7}{7^x}\) IS an integer. So, we have an YES answer.
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Re: M04-26  [#permalink]

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New post 23 Dec 2018, 09:43
I have a question about the approach, why to answer option 2 we have to square? is that a kind of technique?
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New post 23 Dec 2018, 12:05
davidmch1989 wrote:
I have a question about the approach, why to answer option 2 we have to square? is that a kind of technique?


Yes, it's one of the advanced techniques when solving hard modulus questions. Squaring helps to get rid of the absolute values.

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread


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Re M04-26  [#permalink]

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New post 02 Apr 2019, 11:50
I think this is a high-quality question and I don't agree with the explanation. In (1) x is already given as less than equal to 0 or less then equal to 7, so x should be either 0 or 7 and in both cases we get integer as the answer.So, 1 should be sufficient to answer the question
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New post 02 Apr 2019, 21:42
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New post 03 Aug 2019, 10:37
I think this is a high-quality question and I don't agree with the explanation. option 1 is stating clearly its between 0 and 7
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New post 03 Aug 2019, 11:25
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Re: M04-26   [#permalink] 03 Aug 2019, 11:25
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