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aayushagrawal
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it can be resolved on the fly via cyclicity approach by creating a pattern:
Rem[13^1 / 2196] = Rem [13 / 2196] =13
Rem[13^2 / 2196] = Rem [169 / 2196] =49
Rem[13^3 / 2196] = Rem [2197/ 2196] = 1
Rem[13^4 / 2196] = Rem [2197*13 / 2196] = 13

So the cycle is 3 as on the step four it started repeating the same remainder (you can ignore the 4th line for further calculations)

13^36 = 36 / 3 = 12 so it the last 3rd cycle which is 1
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chetan2u
aayushagrawal
What will be the remainder when 13^36 is divided by 2196?

A) 0
B) 1
C)12
D) 2195
E) 5


"Please hit +kudos if you like this post" :-D


In such Qs, best is to get the dividend and divisor in some friendly figures..
\(13^{36} = (13^3)^{12} = 2197^{12} = (2196+1)^{12}\).....
when \((2196+1)^{12}\) is divided by 2196, all terms in the expansion are divisible by 2196 except 1^12, so the remainder will be 1
B


HI Chetan,

why can't we use Cyclicity in here? Is there a particular criteria that has to be fulfilled to use Cyclicity?

TIA
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ajtmatch
chetan2u
aayushagrawal
What will be the remainder when 13^36 is divided by 2196?

A) 0
B) 1
C)12
D) 2195
E) 5


"Please hit +kudos if you like this post" :-D


In such Qs, best is to get the dividend and divisor in some friendly figures..
\(13^{36} = (13^3)^{12} = 2197^{12} = (2196+1)^{12}\).....
when \((2196+1)^{12}\) is divided by 2196, all terms in the expansion are divisible by 2196 except 1^12, so the remainder will be 1
B


HI Chetan,

why can't we use Cyclicity in here? Is there a particular criteria that has to be fulfilled to use Cyclicity?

TIA

hi..
cyclicity generally helps us in knowing the LAST or unit's digit...
so if we are looking for remainder when divided by 10, you can use cyclicity..
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Vidhi96
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When ever you see a situation like this, try to bring numerator as close as denominator so,

(13)^3

=2197/2196 R will always be 1 irrespective the number of power

= (1)^12

= 1 answer
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