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Bunuel
What is the remainder when 6^3 is divided by 8?

(A) 5
(B) 3
(C) 2
(D) 1
(E) 0



Can anyone tell me why it is not possible to come up with ryt answer using cyclicity of 6 ?

we know that unit digit will be 6. if we divide 6 by 8 our remainder will be 6 but actual result will be 0 as \(6^3\) is divisible by 8. thanks.


You can factorise 6 as 2*3. Three 6s means three sets of 2*3. Three 2’s multiply to 8 so you can see it would be divisible.


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Bunuel
What is the remainder when 6^3 is divided by 8?

(A) 5
(B) 3
(C) 2
(D) 1
(E) 0

Note: A number is divisible by 8 if the last two digits are divisible by 8.

6^3=216. Here 16 is divisible by 8 hence, 216 is also divisible.

Answer: (E).
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Bunuel
What is the remainder when 6^3 is divided by 8?

(A) 5
(B) 3
(C) 2
(D) 1
(E) 0

We can re-express 6^3 as (2 x 3)^3, or (2^3)(3^3). Simplifying the expression, we have:

(2^3)(3^3)/2^3 = 3^3, so we have a remainder of zero.

Answer: E
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What is the remainder when 6^3 is divided by 8?

Rem [6^3/8]
=(-2)^3
=-8
=-8+8
=0

correct answer E
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Bunuel
What is the remainder when 6^3 is divided by 8?

(A) 5
(B) 3
(C) 2
(D) 1
(E) 0

Solution:

Since 6^3 = 2^3 * 3^3 and 8= 2^3, we see that 6^3 is divisible by 8. So the remainder is zero.

Answer: E
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We need to find What is the remainder when 6^3 is divided by 8

\(6^3\) = \((2*3)^3\) = \(2^3 *3^3\) = 8 * 27

=> a Multiple of 8

=> Reminder by 8 will be 0

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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