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M07-20

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M07-20  [#permalink]

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New post 16 Sep 2014, 00:35
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

68% (00:58) correct 32% (00:59) wrong based on 53 sessions

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Re M07-20  [#permalink]

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New post 16 Sep 2014, 00:35
Official Solution:

If \(@x= \frac{x^x}{2x^2}-2\), what is the units digit of \(@(@4)\)?

A. 1
B. 3
C. 4
D. 6
E. 8


First of all \(\frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2\), so \(@4=\frac{4^{4-2}}{2}-2=6\);

Next, \(@6=\frac{6^{6-2}}{2}-2=\frac{6^{4}}{2}-2=\frac{6*6^{3}}{2}-2=3*6^3-2\). Now, the units digit of \(6^3\) is 6, thus the units digit of \(3*6^3\) is 8 (\(3*6=18\)), so the units digit of \(3*6^3-2\) is \(8-2=6\).


Answer: D
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Re M07-20  [#permalink]

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New post 23 Dec 2014, 10:05
I think this question is poor and not helpful.
It should be mentioned here that @ is considered as function and not as a digit. Or instead of "http://gmatclub.com/forum/memberlist.php?mode=viewprofile&un= expression f(x) could have been used. Please consider.
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M07-20  [#permalink]

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New post 31 Oct 2015, 21:13
\(4^4\) \(/\) \(2*4^2 - 2\)
\(2^8/2^5 -2\)
\(2^3 = 8 -2 = 6\)
\(6^6/2*6^2\)
\(2^6*3^6 / 2^3 * 3^2 -2\)
\(2^3*3^4 -2\)
\(2^3 = 8\)
\(3^4 = 81\)
81*8 = 648 -2 = 646[/m] so units digit 6.
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Re: M07-20  [#permalink]

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New post 10 Aug 2016, 13:57
We find out that @4 = 6

now while solving @6 : (6^4)/2 - 2


6^1= units digit 6
6^2 = units digit 6
..
..
6^4= units digit 6


Now, when units digit 6 is divided by 2 we get units digit 3.

Units digit 3 -2=1


What did i do wrong?
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Re: M07-20  [#permalink]

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New post 11 Aug 2016, 02:05
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Re: M07-20  [#permalink]

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New post 23 Dec 2018, 07:23
Bunuel wrote:
Official Solution:

If \(@x= \frac{x^x}{2x^2}-2\), what is the units digit of \(@(@4)\)?

A. 1
B. 3
C. 4
D. 6
E. 8


First of all \(\frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2\), so \(@4=\frac{4^{4-2}}{2}-2=6\);

Next, \(@6=\frac{6^{6-2}}{2}-2=\frac{6^{4}}{2}-2=\frac{6*6^{3}}{2}-2=3*6^3-2\). Now, the units digit of \(6^3\) is 6, thus the units digit of \(3*6^3\) is 8 (\(3*6=18\)), so the units digit of \(3*6^3-2\) is \(8-2=6\).


Answer: D
.

how u got this:
\(\frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2\)
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Re: M07-20  [#permalink]

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New post 23 Dec 2018, 12:01
SUNILAA wrote:
Bunuel wrote:
Official Solution:

If \(@x= \frac{x^x}{2x^2}-2\), what is the units digit of \(@(@4)\)?

A. 1
B. 3
C. 4
D. 6
E. 8


First of all \(\frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2\), so \(@4=\frac{4^{4-2}}{2}-2=6\);

Next, \(@6=\frac{6^{6-2}}{2}-2=\frac{6^{4}}{2}-2=\frac{6*6^{3}}{2}-2=3*6^3-2\). Now, the units digit of \(6^3\) is 6, thus the units digit of \(3*6^3\) is 8 (\(3*6=18\)), so the units digit of \(3*6^3-2\) is \(8-2=6\).


Answer: D
.

how u got this:
\(\frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2\)


\(\frac{a^y}{a^y}=a^{x-y}\), so \(\frac{x^x}{x^2}=x^{x-2}\)
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Re: M07-20  [#permalink]

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New post 27 Dec 2018, 08:45
Bunuel wrote:
SUNILAA wrote:
Bunuel wrote:
Official Solution:

If \(@x= \frac{x^x}{2x^2}-2\), what is the units digit of \(@(@4)\)?

A. 1
B. 3
C. 4
D. 6
E. 8


First of all \(\frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2\), so \(@4=\frac{4^{4-2}}{2}-2=6\);

Next, \(@6=\frac{6^{6-2}}{2}-2=\frac{6^{4}}{2}-2=\frac{6*6^{3}}{2}-2=3*6^3-2\). Now, the units digit of \(6^3\) is 6, thus the units digit of \(3*6^3\) is 8 (\(3*6=18\)), so the units digit of \(3*6^3-2\) is \(8-2=6\).


Answer: D
.

how u got this:
\(\frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2\)


\(\frac{a^y}{a^y}=a^{x-y}\), so \(\frac{x^x}{x^2}=x^{x-2}\)

thanks
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Re: M07-20   [#permalink] 27 Dec 2018, 08:45
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