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# M07-20

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Math Expert
Joined: 02 Sep 2009
Posts: 59729

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16 Sep 2014, 00:35
2
4
00:00

Difficulty:

35% (medium)

Question Stats:

66% (01:34) correct 34% (01:30) wrong based on 145 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 59729

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16 Sep 2014, 00:35
1
1
4
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 $$8-2=6$$.

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##### General Discussion
Board of Directors
Joined: 17 Jul 2014
Posts: 2491
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)

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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.
Intern
Joined: 26 May 2014
Posts: 39
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

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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?
Math Expert
Joined: 02 Sep 2009
Posts: 59729

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11 Aug 2016, 02:05
1
devbond wrote:
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?

6/2 = 3, so yes the units digit is 3 but 6^2/2 = 18, the units digit is 8.
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Intern
Joined: 09 Apr 2018
Posts: 34
Location: India
Schools: IIMA PGPX"20
GPA: 3.5

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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 $$8-2=6$$.

.

how u got this:
$$\frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2$$
Math Expert
Joined: 02 Sep 2009
Posts: 59729

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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 $$8-2=6$$.

.

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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Intern
Joined: 09 Apr 2018
Posts: 34
Location: India
Schools: IIMA PGPX"20
GPA: 3.5

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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 $$8-2=6$$.

.

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
Intern
Joined: 06 Jun 2017
Posts: 8

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02 Sep 2019, 07:27
I wrote 6^4/2 -2 as (6^4 - 4) /2

Hence, unit digit of 6^4 is 6 : 6-4 =2 and 2/2=1 . What did I do wrong?

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 59729

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02 Sep 2019, 07:49
Sumit1703@4 wrote:
I wrote 6^4/2 -2 as (6^4 - 4) /2

Hence, unit digit of 6^4 is 6 : 6-4 =2 and 2/2=1 . What did I do wrong?

Posted from my mobile device

The point is that if the units digits of a number is 2, the units digit of that number divided by 2 won't always be 1, it could also be 6. For example, 22/2 = 11 but 12/2 = 6.
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Intern
Joined: 11 Nov 2019
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21 Nov 2019, 10:07
Why do we solve for @6 when @4 was already solved to be 6? Thank you.

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 59729

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22 Nov 2019, 01:11
Django86 wrote:
Why do we solve for @6 when @4 was already solved to be 6? Thank you.

Posted from my mobile device

Notice that there are TWO symbols representing function $$@(@4)$$.
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Re: M07-20   [#permalink] 22 Nov 2019, 01:11
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# M07-20

Moderators: chetan2u, Bunuel