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Re M0426
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15 Sep 2014, 23:23
Official Solution:Is \(\frac{7^7}{7^x}\) an integer? \(\frac{7^7}{7^x} = 7^{7x}\), 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: M0426
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26 Nov 2014, 12:20
Bunuel wrote: Official Solution:
\(\frac{7^7}{7^x} = 7^{7x}\), 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: M0426
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27 Nov 2014, 05:49
prashd wrote: Bunuel wrote: Official Solution:
\(\frac{7^7}{7^x} = 7^{7x}\), 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: M0426
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28 Nov 2014, 06:31
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: M0426
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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



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Re: M0426
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07 Oct 2016, 16:21
Bunuel wrote: prashd wrote: Bunuel wrote: Official Solution:
\(\frac{7^7}{7^x} = 7^{7x}\), 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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08 Oct 2016, 02:07



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Re: M0426
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25 Oct 2018, 22:27
I think this is a highquality 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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25 Oct 2018, 23:00



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Re: M0426
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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?



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23 Dec 2018, 11:05










