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# Find the remainder when 25^18 is divided by 9

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Joined: 20 Jul 2017
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Find the remainder when 25^18 is divided by 9  [#permalink]

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02 Apr 2020, 04:06
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62% (01:47) correct 38% (01:39) wrong based on 60 sessions

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Find the remainder when $$25^{18}$$ is divided by 9.

A. 1
B. 2
C. 3
D. 5
E. 6
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Find the remainder when 25^18 is divided by 9  [#permalink]

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02 Apr 2020, 04:13
Dillesh4096 wrote:
Find the remainder when $$25^{18}$$ is divided by 9.

A. 1
B. 2
C. 3
D. 5
E. 6

$$R[\frac{25^{18}}{9}] = R[\frac{(27-2)^{18}}{9}] = R[\frac{(-2)^{18}}{9}] = R[\frac{(2)^{18}}{9}]$$

$$R[\frac{(2)^{3}}{9}] = 8 or -1$$ (negative remainder)

CONCEPT: Remainder can be written in both +ve and -ve form for simplification. When divisor is 9 and remainder is 7 then the number has either 7 in excess or number is 2 short to be divisible by 9 hence
Remainder (+7) = Remainder (-2) when divisor is 9

Taking exponent 6 both sides

$$R[\frac{(2)^{18}}{9}] = (-1)^6 = 1$$

The concept video of remainder theorem is as follows:

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Re: Find the remainder when 25^18 is divided by 9  [#permalink]

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02 Apr 2020, 06:14
METHOD I : Euler no of 9 = 9(1-1/3) = 6
rem [(25^18) /9]
= rem [(25^0)/9] [ 18 is divisible by 6 ]
= rem [1/9]
= 1

METHOD II : rem [(25^18 )/9]
=rem [ (-2)^18 /9] [ rem (25/9) = -2 ]
=rem [ (+2)^18 /9]
=rem [ (2^6) * ( 2^6 )* (2^6 )/9]
=rem [ 64*64*64 /9]
=rem [1*1*1 /9]
=1

METHOD III : rem [(25^18 )/9]
= rem [ (27-2)^18/9] [ using binomial theorem ]
=rem [ (-2)^18 /9] [ considering only last term since rest all terms will be divisible by 9 ]
=rem [ (+2)^18 /9]
=rem [ (2^6) * ( 2^6 )* (2^6 )/9]
=rem [ 64*64*64 /9]
=rem [1*1*1 /9]
=1

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Joined: 12 Mar 2019
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Re: Find the remainder when 25^18 is divided by 9  [#permalink]

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27 May 2020, 02:50
preetamsaha wrote:
METHOD I : Euler no of 9 = 9(1-1/3) = 6
rem [(25^18) /9]
= rem [(25^0)/9] [ 18 is divisible by 6 ]
= rem [1/9]
= 1

METHOD II : rem [(25^18 )/9]
=rem [ (-2)^18 /9] [ rem (25/9) = -2 ]
=rem [ (+2)^18 /9]
=rem [ (2^6) * ( 2^6 )* (2^6 )/9]
=rem [ 64*64*64 /9]
=rem [1*1*1 /9]
=1

METHOD III : rem [(25^18 )/9]
= rem [ (27-2)^18/9] [ using binomial theorem ]
=rem [ (-2)^18 /9] [ considering only last term since rest all terms will be divisible by 9 ]
=rem [ (+2)^18 /9]
=rem [ (2^6) * ( 2^6 )* (2^6 )/9]
=rem [ 64*64*64 /9]
=rem [1*1*1 /9]
=1

hi, can you tell about euler theorem, What i know is
E(z)=z∗[1–1/P]∗[1–1/Q], where P and Q are prime factors of denominator
do we need 1 factor or 2 prime factors. In case we don't have 2 prime factors, Is 1 prime factor ok
Re: Find the remainder when 25^18 is divided by 9   [#permalink] 27 May 2020, 02:50