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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 Source: GMAT Club Tests  hardest GMAT questions OA: Based on the formula, the value of $4$ is which is or . However, in the numerator will be a large number with the last digit being 4  you can get it just multiplying and getting 4 as a last digit for every 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 " 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: m0781002.html#p1233091



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Re: M07#20 [#permalink]
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18 Jul 2009, 16:26
DFG5150 wrote: If , what is the last digit of $($4$)$? 8 6 4 2 1 OA: Based on the formula, the value of $4$ is which is or . However, in the numerator will be a large number with the last digit being 4  you can get it just multiplying and getting 4 as a last digit for every 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 " 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]
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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 Source: GMAT Club Tests  hardest GMAT questions OA: Based on the formula, the value of $4$ is which is or . However, in the numerator will be a large number with the last digit being 4  you can get it just multiplying and getting 4 as a last digit for every 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 and getting 4 as a last digit for every" Thanks in advance.
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Re: M07#20 [#permalink]
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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



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Re: M07#20 [#permalink]
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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
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Re: M07#20 [#permalink]
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31 May 2010, 10:57
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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) = 102 4 x 12 8last digit from: 4x8 = 3 2 that is 2. OA = D
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Re: M07#20 [#permalink]
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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 Source: GMAT Club Tests  hardest GMAT questions OA: Based on the formula, the value of $4$ is which is or . However, in the numerator will be a large number with the last digit being 4  you can get it just multiplying and getting 4 as a last digit for every 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 " 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]
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31 May 2010, 20:05
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]
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13 Jun 2010, 05:09
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]
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28 Jun 2010, 03:50
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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]
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02 Jun 2011, 05:00
Let us start from the bracket : @4@ = 4^2/2= 16/2 = 8 Then @8@ = 8^(82)/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]
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02 Jun 2011, 05:50
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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]
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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.



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Re: M07#20 [#permalink]
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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..
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Re: M07#20 [#permalink]
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02 Jun 2011, 23:49
Simple and clear explanation srini88



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Re: M07#20 [#permalink]
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15 Mar 2012, 10:01
subhashghosh wrote: Let us start from the bracket :
@4@ = 4^2/2= 16/2 = 8
Then @8@ = 8^(82)/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]
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06 Jun 2013, 05:27



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Re: M07#20 [#permalink]
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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^(82)/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.










