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# M04-26

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:23
4
12
00:00

Difficulty:

55% (hard)

Question Stats:

50% (00:47) correct 50% (00:38) wrong based on 271 sessions

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Is $$\frac{7^7}{7^x}$$ an integer?

(1) $$0 \le x \le 7$$

(2) $$|x| = x^2$$

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Joined: 02 Sep 2009
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15 Sep 2014, 23:23
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.

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26 Nov 2014, 12:20
1
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).

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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27 Nov 2014, 05:49
1
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).

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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28 Nov 2014, 06:31
2
1
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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24 Jul 2016, 06:47
Just a small doubt
st 2) three values of x are derived... -1 , 0 ,1
so three different values... and stiff st sufficient?
thanks
Math Expert
Joined: 02 Sep 2009
Posts: 52463

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24 Jul 2016, 07:24
Celestial09 wrote:
Just a small doubt
st 2) three values of x are derived... -1 , 0 ,1
so three different values... and stiff st sufficient?
thanks

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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Joined: 22 Jun 2016
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07 Oct 2016, 16:21
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).

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.

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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Joined: 02 Sep 2009
Posts: 52463

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08 Oct 2016, 02:07
jonmarrow wrote:

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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25 Oct 2018, 22:27
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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Joined: 02 Sep 2009
Posts: 52463

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25 Oct 2018, 23:00
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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Joined: 03 Oct 2018
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23 Dec 2018, 08:43
I have a question about the approach, why to answer option 2 we have to square? is that a kind of technique?
Math Expert
Joined: 02 Sep 2009
Posts: 52463

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23 Dec 2018, 11:05
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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# M04-26

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