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sriamlan
What is the remainder when \(11^{452}\) is divided by 6.

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Experts my doubt here is -
The cyclicity of power of 11 when divided by 6 is 5,4,3,2,1,0

So if we divide 452 by 6 remainder is 2

So as 2 is the remainder as per the cyclicity should not 4 be the answer instead of 1?
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sriamlan
What is the remainder when \(11^{452}\) is divided by 6.

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

\(\frac{11}{6}=\frac{5}{6} \implies \frac{11^{452}}{6}=\frac{5^{452}}{6}=\frac{(-1)^{452}}{6}=\frac{1}{6}\)

The answer is A.
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sriamlan
What is the remainder when \(11^{452}\) is divided by 6.

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

\(\frac{11}{6}\) = Remainder \(5\)
\(\frac{11^2}{6}\) = Remainder \(1\)

Now, \(11^{452}\) = \(11^{2*226}\)

\(\frac{11^{2*226}}{6}\) will leave a remainder 1

Thus, the correct answer will be (A) 1
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sriamlan
sriamlan
What is the remainder when \(11^{452}\) is divided by 6.

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Experts my doubt here is -
The cyclicity of power of 11 when divided by 6 is 5,4,3,2,1,0

So if we divide 452 by 6 remainder is 2

So as 2 is the remainder as per the cyclicity should not 4 be the answer instead of 1?


Not sure if i can clear your doubt. You can look at the problem like when 11/6 then remainder is 5, when 11^2/6 then remainder is 1.
Now you have 11^452 which is 11^even power so remainder will be 1.
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6 = 3 * 2
11^452 = (11^256)*(11^256)
a^(p-1)/p = 1 mod p

11^(3-1)/3 = 11^2 = 1 mod 3
11^256= 11^(2*128) = 1 mod 3
(11^256)/2 = 1 mod 2

together 1
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Rem|11^452/6| is equivalent to Rem|5^452/6|

Now rem|5^2/6| = 1

So Rem|5^452/6| = Rem|(5^2)^226/6|

=> 1^226 = 1

Final Remainder = 1
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sriamlan
What is the remainder when \(11^{452}\) is divided by 6.

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

This problem is pretty simple. because if we divide 11 with 6 , we get remainder as -1 , final remainder will be (-1)*452 , which is 1. if exponent had been odd say 1999, the remainder would have been -1. you can ask what if dividend doesnt go by divisor without 1 or -1 as remainder, The thing is first we try use the exponent to make the divisor to take form of (divisor*K + 1) or (Divisor*K - 1), easy and medium questions fall to this category. However for hard questions we need to use cyclicity.
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Hello!

Is it possible to do the following?

11-6= -5 Negative R

6-5= 1 R

Kind regards!
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sriamlan
What is the remainder when \(11^{452}\) is divided by 6.

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5


Hi..

There are two ways you could do this..

1) cylicity..
11 leaves a remainder of 5.
\(11^2\) or 121 leaves a remainder of 1.
\(11^3=1331\) will leave a remainder of 5..
So we have a cylicity of 5,1,5,1...
Thus every odd power will give 5 as remainder and every even power will give 1 as remainder.
452 is even, so remainder is 1.

2) binomial expansion
\(11^{452}=(12-1)^{452}\)
The expansion will have all terms containing 12 except the last that will be \(12^0*(-1)^{452}\)=1..
So remainder will be 1

A
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We need to find the remainder when \(11^{452}\) is divided by 6?

We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(6) and a big number, other part is a small number.

=> \(11^{452}\) = \((12-1)^{452}\)

Watch this video to MASTER BINOMIAL Theorem

Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 6
=> All terms except the last term will give remainder of 0 when divided by 6
=> Problem is reduced to what is the remainder when the last term (i.e. 452C452 * 12^0 * (-1)^452) is divided by 6
=> Remainder of 1 is divided by 6
=> Remainder of 1

So, Answer will be A
Hope it Helps!

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

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