It is currently 28 Jun 2017, 23:19

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M07#20

Author Message
Intern
Joined: 06 May 2008
Posts: 28

### Show Tags

17 Jul 2009, 08:21
1
KUDOS
1
This post was
BOOKMARKED
If $$\Large{@}\large{Y}\Large{@} = \frac{Y^Y}{2Y^2}$$ , what is the last digit of $$\Large{@}\large{(}\Large{@}\large{4}\Large{@}\large{)}\Large{@}$$ ?

(A) 8
(B) 6
(C) 4
(D) 2
(E) 1

Source: GMAT Club Tests - hardest GMAT questions

[Reveal] Spoiler:
OA:
Based on the formula, the value of $4$ is $\frac{4^4}{2*4^2}$ which is $\frac{4^2}{2}$ or $\frac{16}{2} = 8$ .

However, in $\frac{8^6}{2}$ the numerator will be a large number with the last digit being 4 - you can get it just multiplying $8^2*8^2*8^2$ and getting 4 as a last digit for every $4*4*4$ returns 4 as a last digit. So, 4 divided by 2 is 2.

End of OA.
-----------------------------------------------------------------------------------------------------------------

My problem is that when the OA says "$4*4*4$ returns 4 as a last digit. So, 4 divided by 2 is 2" it assumes that the last digit of a quotient is the last digit of the division between the last digits of the numerator and the denominator which is not always true... in this example it worked since 64/2=32 and 4/2=2... but what about 14/2?? The result is 7, not 2....

REVISED VERSION OF THIS QUESTION IS HERE: m07-81002.html#p1233091
Founder
Joined: 04 Dec 2002
Posts: 15153
Location: United States (WA)
GMAT 1: 750 Q49 V42

### Show Tags

18 Jul 2009, 16:26
DFG5150 wrote:
If$Y = \frac{Y^Y}{2Y^2}$ , what is the last digit of $($4$)$?

8
6
4
2
1

OA:
Based on the formula, the value of $4$ is $\frac{4^4}{2*4^2}$ which is $\frac{4^2}{2}$ or $\frac{16}{2} = 8$ .

However, in $\frac{8^6}{2}$ the numerator will be a large number with the last digit being 4 - you can get it just multiplying $8^2*8^2*8^2$ and getting 4 as a last digit for every $4*4*4$ returns 4 as a last digit. So, 4 divided by 2 is 2.

End of OA.
-----------------------------------------------------------------------------------------------------------------

My problem is that when the OA says "$4*4*4$ returns 4 as a last digit. So, 4 divided by 2 is 2" it assumes that the last digit of a quotient is the last digit of the division between the last digits of the numerator and the denominator which is not always true... in this example it worked since 64/2=32 and 4/2=2... but what about 14/2?? The result is 7, not 2....

I don't think that's relevant. In this question, you are specifically asked for the last digit of $($4$)$.
_________________

Founder of GMAT Club

US News Rankings progression - last 10 years in a snapshot - New!
Just starting out with GMAT? Start here...
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

Manager
Joined: 24 Jan 2010
Posts: 156
Location: India
Schools: ISB

### Show Tags

31 May 2010, 05:34
DFG5150 wrote:
If $$\Large{@}\large{Y}\Large{@} = \frac{Y^Y}{2Y^2}$$ , what is the last digit of $$\Large{@}\large{(}\Large{@}\large{4}\Large{@}\large{)}\Large{@}$$ ?

(A) 8
(B) 6
(C) 4
(D) 2
(E) 1

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

OA:
Based on the formula, the value of $4$ is $\frac{4^4}{2*4^2}$ which is $\frac{4^2}{2}$ or $\frac{16}{2} = 8$ .

However, in $\frac{8^6}{2}$ the numerator will be a large number with the last digit being 4 - you can get it just multiplying $8^2*8^2*8^2$ and getting 4 as a last digit for every $4*4*4$ returns 4 as a last digit. So, 4 divided by 2 is 2.

End of OA.

Hello BB,

I got the same answer but with a longer method.

can you please explain "the numerator will be a large number with the last digit being 4 - you can get it just multiplying $8^2*8^2*8^2$ and getting 4 as a last digit for every"

_________________

_________________
If you like my post, consider giving me a kudos. THANKS!

Manager
Joined: 18 Jul 2009
Posts: 51

### Show Tags

31 May 2010, 07:46
(8^6)/2= (8^5 * 2^3)/2 = (8^5 * 4)=

we know last digit 8^5 = 8 ^ 1= 8

so last digit of (8^5 * 4) is 2
Manager
Affiliations: NCC,SAE,YHIA
Joined: 04 May 2010
Posts: 51
Location: Mumbai , India
WE 1: 3 years international sales & mktg-projects

### Show Tags

31 May 2010, 10:21
(8^6)/2= (2^18)/2
=2^17
Remainder left when 17 is divided by 4 :- 17/4 = 1
Therefore last digit shall be 2^1=2
_________________

Sun Tzu-Victorious warriors win first and then go to war, while defeated warriors go to war first and then seek to win.

Director
Joined: 21 Dec 2009
Posts: 583
Concentration: Entrepreneurship, Finance

### Show Tags

31 May 2010, 10:57
1
KUDOS
My method, i think is rather lengthy:
(4^4)/2(4)^2 = 16/2 = 8
(8^8)/(2(8^2) = 8^6/2 = 2^17
2^17 = (2^10)(2^7) = 1024 x 128
last digit -from: 4x8 = 32
that is 2. OA = D
_________________

KUDOS me if you feel my contribution has helped you.

Director
Joined: 03 May 2007
Posts: 872
Schools: University of Chicago, Wharton School

### Show Tags

31 May 2010, 11:18
DFG5150 wrote:
If $$\Large{@}\large{Y}\Large{@} = \frac{Y^Y}{2Y^2}$$ , what is the last digit of $$\Large{@}\large{(}\Large{@}\large{4}\Large{@}\large{)}\Large{@}$$ ?

(A) 8
(B) 6
(C) 4
(D) 2
(E) 1

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

OA:
Based on the formula, the value of $4$ is $\frac{4^4}{2*4^2}$ which is $\frac{4^2}{2}$ or $\frac{16}{2} = 8$ .

However, in $\frac{8^6}{2}$ the numerator will be a large number with the last digit being 4 - you can get it just multiplying $8^2*8^2*8^2$ and getting 4 as a last digit for every $4*4*4$ returns 4 as a last digit. So, 4 divided by 2 is 2.

End of OA.
-----------------------------------------------------------------------------------------------------------------

My problem is that when the OA says "$4*4*4$ returns 4 as a last digit. So, 4 divided by 2 is 2" it assumes that the last digit of a quotient is the last digit of the division between the last digits of the numerator and the denominator which is not always true... in this example it worked since 64/2=32 and 4/2=2... but what about 14/2?? The result is 7, not 2....

Consider this way:- (8^k)/2 never has a factor i.e. odd except for 1. So in the given expression, 1, and 7 are not possible. Now remain - 2, 4, 6 and 8. (8^k) has never 2 in unit digit when k is even so 6 is also not possible. Similarly 8 is also not possible when k is not equal to 2^2n where n is an integer. After a but further simplification, 2 remains between 2 and 4.

The explanation is good enough..
Manager
Joined: 04 Dec 2009
Posts: 70
Location: INDIA

### Show Tags

31 May 2010, 20:05
Ans is 2

multiplication of 4*4*4 =64 last digit is 4/2 so ANS:D
_________________

MBA (Mind , Body and Attitude )

Intern
Joined: 17 May 2010
Posts: 19

### Show Tags

13 Jun 2010, 05:09
i also came up with 2, but involved quite a few calculation. Thanks with 8(square)*8(square)
Intern
Joined: 08 Jun 2010
Posts: 31
GMAT 1: 690 Q48 V36
GPA: 3.59
WE: Accounting (Accounting)

### Show Tags

28 Jun 2010, 03:50
1
KUDOS
I have a question what is that all @ signs? I do not get questions? What is being asked?
SVP
Joined: 16 Nov 2010
Posts: 1663
Location: United States (IN)
Concentration: Strategy, Technology

### Show Tags

02 Jun 2011, 05:00
Let us start from the bracket :

@4@ = 4^2/2= 16/2 = 8

Then @8@ = 8^(8-2)/2 = 8^6/2 = 2^18/2 = 2^17

Now 2 has a cyclicity of 4, so 17/4 = Remainder 1

=> Last digit of 2^17 = 2^1 = 2

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 01 Jun 2011
Posts: 7

### Show Tags

02 Jun 2011, 05:50
1
KUDOS
8^6/2

Just ignore the tens + digits

8^2 = singles unit: 4
8^3= 4x8= 2 singles digit
8^4= 2x8= 6 singles digit
8^5= 6x8= 8 singles digit
8^6= 8x8= 4 singles

Intern
Joined: 24 May 2010
Posts: 46

### Show Tags

02 Jun 2011, 06:12
samatace wrote:
I have a question what is that all @ signs? I do not get questions? What is being asked?

Don't be more worried with '@'. It could be any other symbol also, in place of @. All we need to be focusing is, what to be substituted in place of 'Y' in the given equation. Given a equation for @Y@ = $$Y^Y/2Y^2$$, they asking for @(@4@)@. So, first you calculate @4@, and second, calculate the same for the result of @4@.

1. @4@ = 8
2. @8@ = some number ending with '2'. We dont need to spend too much time in calculating the whole #, as question is looking only for the last digit.

Hope you understood.
Manager
Joined: 28 Feb 2011
Posts: 85

### Show Tags

02 Jun 2011, 18:08
Took a slightly different approach at the end.

@4@ = 4^2/2= 16/2 = 8

@8@ = (8^6)/2
> (8^5) x 4

Now, the cyclicity of 8 is 4..

so xxxx8 times 4..last digit 2..
_________________

Fight till you succeed like a gladiator..or doom your life which can be invincible

Intern
Joined: 02 Jun 2011
Posts: 5

### Show Tags

02 Jun 2011, 23:49
Simple and clear explanation srini88
Intern
Joined: 21 Feb 2012
Posts: 12
Location: Chile

### Show Tags

15 Mar 2012, 10:01
subhashghosh wrote:
Let us start from the bracket :

@4@ = 4^2/2= 16/2 = 8

Then @8@ = 8^(8-2)/2 = 8^6/2 = 2^18/2 = 2^17

Now 2 has a cyclicity of 4, so 17/4 = Remainder 1

=> Last digit of 2^17 = 2^1 = 2

I think this is a good approach. It is worth to remember the clyclicity of some numbers. For instance, numbers 2, 3 and 7 has the same cyclicity: 4. That is, the units digit of the powers of 2, 3 and 7 are repeated after every 4 powers. Remembering this, you could save a couple of seconds during the test.
_________________

Francisco

Math Expert
Joined: 02 Sep 2009
Posts: 39755

### Show Tags

06 Jun 2013, 05:27
1
KUDOS
Expert's post
BELOW IS REVISED VERSION OF THIS QUESTION:

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.

_________________
Intern
Joined: 30 May 2013
Posts: 13

### Show Tags

06 Jun 2013, 09:30
Hello, this can be approached by observing how the units digit in powers of 2 repeat in groups of 4.

@4@ = 4^2/2= 16/2 = 8

Then @8@ = 8^(8-2)/2 = 8^6/2 = 2^18/2 = 2^17

Now for powers of 2, the units digit goes like : 2, 4, 8, 6, 2, 4...

Thus 17th power of 2 will have 4 groups of (2, 4, 8, 6) and 17/4 gives a remainder of 1. => Units digit of 2^17 = 2^1 = 2

Re: M07#20   [#permalink] 06 Jun 2013, 09:30
Display posts from previous: Sort by

# M07#20

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.