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M07#20

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M07#20 [#permalink] New post 17 Jul 2009, 07:21
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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

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[Reveal] Spoiler:
OA:
Based on the formula, the value of $4$ is Image which is Image or Image .

However, in Image the numerator will be a large number with the last digit being 4 - you can get it just multiplying Image and getting 4 as a last digit for every Image returns 4 as a last digit. So, 4 divided by 2 is 2.

The correct answer is D.

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

My problem is that when the OA says "Image 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
[Reveal] Spoiler: OA
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Re: M07#20 [#permalink] New post 18 Jul 2009, 15:26
Expert's post
DFG5150 wrote:
IfImage , what is the last digit of $($4$)$?

8
6
4
2
1


OA:
Based on the formula, the value of $4$ is Image which is Image or Image .

However, in Image the numerator will be a large number with the last digit being 4 - you can get it just multiplying Image and getting 4 as a last digit for every Image returns 4 as a last digit. So, 4 divided by 2 is 2.

The correct answer is D.

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

My problem is that when the OA says "Image 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 $($4$)$.
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Re: M07#20 [#permalink] New post 31 May 2010, 04:34
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 Image which is Image or Image .

However, in Image the numerator will be a large number with the last digit being 4 - you can get it just multiplying Image and getting 4 as a last digit for every Image returns 4 as a last digit. So, 4 divided by 2 is 2.

The correct answer is D.

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 Image and getting 4 as a last digit for every"

Thanks in advance.
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Re: M07#20 [#permalink] New post 31 May 2010, 06:46
(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
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Re: M07#20 [#permalink] New post 31 May 2010, 09:21
(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
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Re: M07#20 [#permalink] New post 31 May 2010, 09:57
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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
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Re: M07#20 [#permalink] New post 31 May 2010, 10:18
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 Image which is Image or Image .

However, in Image the numerator will be a large number with the last digit being 4 - you can get it just multiplying Image and getting 4 as a last digit for every Image returns 4 as a last digit. So, 4 divided by 2 is 2.

The correct answer is D.

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

My problem is that when the OA says "Image 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..
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Re: M07#20 [#permalink] New post 31 May 2010, 19:05
Ans is 2

multiplication of 4*4*4 =64 last digit is 4/2 so ANS:D
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Re: M07#20 [#permalink] New post 13 Jun 2010, 04:09
i also came up with 2, but involved quite a few calculation. Thanks with 8(square)*8(square)
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Re: M07#20 [#permalink] New post 28 Jun 2010, 02:50
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I have a question what is that all @ signs? I do not get questions? What is being asked?
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Re: M07#20 [#permalink] New post 02 Jun 2011, 04:00
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

Answer - D
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Re: M07#20 [#permalink] New post 02 Jun 2011, 04:50
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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

4/2= 2 Answer D
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Re: M07#20 [#permalink] New post 02 Jun 2011, 05:12
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.
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Re: M07#20 [#permalink] New post 02 Jun 2011, 17:08
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..
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Re: M07#20 [#permalink] New post 02 Jun 2011, 22:49
Simple and clear explanation srini88 :)
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Re: M07#20 [#permalink] New post 15 Mar 2012, 09:01
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

Answer - D

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.
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Re: M07#20 [#permalink] New post 06 Jun 2013, 04:27
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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.

Answer: D.
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Re: M07#20 [#permalink] New post 06 Jun 2013, 08:30
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

So the answer is D. :)
Re: M07#20   [#permalink] 06 Jun 2013, 08:30
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