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GMATinsight sir can you shed some light on this

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GMATinsight

Cyclicity of 4 is 2 ( 4,16,64...)

Means units digit of 4^96 is 6

And on division by 6 it should leave a reminder zero.

But in this case the answer is 4.

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s111
GMATinsight

Cyclicity of 4 is 2 ( 4,16,64...)

Means units digit of 4^96 is 6

And on division by 6 it should leave a reminder zero.

But in this case the answer is 4.

Posted from my mobile device

You are seeing the remainder of units digit not whole value.. which is why you are not getting 4 as answer
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BUT we always see units digit only to solve for reminders? please correct me here if i am wrong

487^191 divided by 5
cyclicity of 7 is 7,9,3,1 = 4
dividing 191/4 we get 47 pairs so units digit is 3

so reminder is 3
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BUT we always see units digit only to solve for reminders? please correct me here if i am wrong

487^191 divided by 5
cyclicity of 7 is 7,9,3,1 = 4
dividing 191/4 we get 47 pairs so units digit is 3

so reminder is 3

NO.. PLEASE CHECK WHAT WOULD BE REMAINDER WHEN 4^2 IS DIVIDED BY 6 .. WILL YOU GET 0 OR 4..

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Bunuel
What is the remainder when 4^96 is divided by 6?


A. 0

B. 1

C. 2

D. 3

E. 4

Archit3110

This is NOT a question of unit digit calculation. this instead is a question of cyclicity of remainders

\(4^1\) divided by 6 leaves remainder = 4
\(4^2\) divided by 6 leaves remainder = 4
\(4^3\) divided by 6 leaves remainder = 4
\(4^4\) divided by 6 leaves remainder = 4

i.e. remainder is always 4 irrespective of any exponent of 4 when the number \(4^x\) where m ≥1 is divided by 6

Answer: Option E

I hope this helps!!!
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Archit3110
s111
BUT we always see units digit only to solve for reminders? please correct me here if i am wrong

487^191 divided by 5
cyclicity of 7 is 7,9,3,1 = 4
dividing 191/4 we get 47 pairs so units digit is 3

so reminder is 3

NO.. PLEASE CHECK WHAT WOULD BE REMAINDER WHEN 4^2 IS DIVIDED BY 6 .. WILL YOU GET 0 OR 4..

Posted from my mobile device


Do NOT divide unit digit of number by 6 to calculate the remainder. Divide the entire number by 6 and then see the remainder.

Unit digit is reponsible for remainder only when the divisior is either 5 or 10
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Bunuel
What is the remainder when 4^96 is divided by 6?


A. 0

B. 1

C. 2

D. 3

E. 4

Archit3110

This is NOT a question of unit digit calculation. this instead is a question of cyclicity of remainders

\(4^1\) divided by 6 leaves remainder = 4
\(4^2\) divided by 6 leaves remainder = 4
\(4^3\) divided by 6 leaves remainder = 4
\(4^4\) divided by 6 leaves remainder = 4

i.e. remainder is always 4 irrespective of any exponent of 4 when the number \(4^x\) where m ≥1 is divided by 6

Answer: Option E

I hope this helps!!!

GMATinsight , I had understood the question , as rightly pointed out by you this is not a cyclcity question, I actually used cyclcity to determine the unit value of 4^96..but nonetheless it isn't of much use as remainder would always be' 4 'when divided by 6

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I did this way:-

(4^96)/6= (4X4^95)/6

Above can be reduced to this, 2X (4^95)/3 {Keep a note we have cancelled 2 from neumerator & Denomenator so final result to be mutliplied by this}

Now, 2X 4^95/3 remainder= 2X (Rem. 4/3)^95=2 X (1)^95=2

Final Remainder= 2X2=4
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GMATinsight
Hi...am wondering if the following is correct:

4^96 = 2^192 = (3 - 1)^192
Now:

(3-1)^192/6 --> the '3' raised to any power will give a remainder of 3 when divided by 6 AND '1' will give a remainder of 1 when divided by 6, but since the minus sign has an even power we get:
r=3+1=4

your comments will be appreciated
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Mansoor50
GMATinsight
Hi...am wondering if the following is correct:

4^96 = 2^192 = (3 - 1)^192
Now:

(3-1)^192/6 --> the '3' raised to any power will give a remainder of 3 when divided by 6 AND '1' will give a remainder of 1 when divided by 6, but since the minus sign has an even power we get:
r=3+1=4

your comments will be appreciated

Mansoor50

Your reasoning is 100% correct here... :)
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Mansoor50
GMATinsight
Hi...am wondering if the following is correct:

4^96 = 2^192 = (3 - 1)^192
Now:

(3-1)^192/6 --> the '3' raised to any power will give a remainder of 3 when divided by 6 AND '1' will give a remainder of 1 when divided by 6, but since the minus sign has an even power we get:
r=3+1=4

your comments will be appreciated

Hello Mansoor50
I solved it the same way you did, with 1 exception. I think you created an extra step by writing 4^96 as 2^192, and further writing 2^192 as (3-1)^192.

Instead, we can write 4^96 as (3+1)^96.
Then remainder for 3/6 = 3
Remainder for 1/6 = 1
Final remainder = 3+1 = 4

Just shared this solution because I think it saves a little bit more time to do it this way. :)
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Is this 700 level question?
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Is this 700 level question?

You can check difficulty level of a question along with the stats on it in the first post. For this question Difficulty = 700 Level. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.
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\(4^{96} = (2^2)^{96}= 2^{192}\)

\(2^{4*48} = 2^4 = 16 \)

=> 16 divided by 6 will give the remainder 4

Answer E
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We need to find the remainder when \(4^{96}\) is divided by 6?

\(4^{96}\) = \((2^2)^{96}\) = \(2^{2*96}\) = \(2^{192}\)

To solve this problem we need to find the cycle of remainder of power of 2 when divided by 6

Remainder of \(2^1\) (=2) by 6 = 2
Remainder of \(2^2\) (=4) by 6 = 4
Remainder of \(2^3\) (=8) by 6 = 2
Remainder of \(2^4\) (=16) by 6 = 4


=> Cycle is 2

=> Odd Power remainder is 2
=> Even power remainder is 4
=> Remainder of \(2^{192}\) is divided by 6 = 4


So, Answer will be E
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

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How can we solve this ques using binomial theorem?
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