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M04-26

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M04-26  [#permalink]

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New post 15 Sep 2014, 23:23
4
12
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

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

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Re M04-26  [#permalink]

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New post 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.


Answer: B
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Re: M04-26  [#permalink]

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New post 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).


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, 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).


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, 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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Re: M04-26  [#permalink]

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New post 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
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Re: M04-26  [#permalink]

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New post 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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Re: M04-26  [#permalink]

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New post 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).


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, 02:07
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, 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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Re: M04-26  [#permalink]

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New post 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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Re: M04-26  [#permalink]

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New post 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?
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Re: M04-26  [#permalink]

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New post 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



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Re: M04-26 &nbs [#permalink] 23 Dec 2018, 11:05
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