What is the remainder when ((11^12)^13)^14 is divided by9?

Question

What is the remainder when  ((11^{12})^{13})^{14}  is divided by9?

A. 1
B. 2
C. 3
D. 5
E. 7

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chetan2u · Accepted Answer

Let us expand the term  ((11^{12})^{13})^{14} 
 ((11^{12})^{13})^{14} =  ((9+2)^{12})^{13})^{14} 
 When we expand the term using binomial expansion, all terms except the last will be divisible by 9

((2^{12})^{13})^{14}

((2^3)^4)^{13})^{14}

((8)^4)^{13})^{14}

((9-1)^4)^{13})^{14} 
 Again only last term will not be divisible by 9, when the term is expanded

((-1)^4)^{13})^{14}

(1^{13})^{14}=1

So remainder =1 .

We can also find pattern
11 leaves a remainder 2
 11^2.....2^2 
 11^3.....2^3=8 
 11^4......2^4=16=9+7 
 11^5......2^5=32=27+5 
 11^6......2^6=64=63+1 
 11^7......2^7=128=126+2 
 So pattern repeats after every 6 terms  => 2,4,8,7,5,1,2,4,8....
Now the power 12 is multiple of 6, so  11^{12}  will leave 1 as the remainder.

A

danbrown9445 · Answer

((11^{12})^{13})^{14} 
 ((121^{6})^{13})^{14} 
 (((117+4)^{6})^{13})^{14}

117 = 13*9 so it wont leave any remainder when divided by 9

((4^{6})^{13})^{14} 
 ((64^{2})^{13})^{14} 
 (((63+1)^{2})^{13})^{14}

(((1)^{2})^{13})^{14} 
 =1

BrushMyQuant · Answer

We need to find the remainder of  ((11^{12})^{13})^{14}  by 9 ?

((11^{12})^{13})^{14}  =  11^{(12*13*14)}  =  11^{2184}

We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(9) and a big number, other part is a small number.

=>  11^ { 2184}  =  (9 + 2)^ { 2184}

Watch this video to  MASTER BINOMIAL Theorem

Now, when we expand this expression then all the terms except the last term will be a multiple of 9. 
=> All terms except the last term will give 0 as remainder then divided by 9
=> Problem is reduced to what is the remainder when the last term is divided by 9
=> What is the remainder when  2184C2184 * 9^0 * 2^ { 2184}  is divided by 9 = Remainder of  2^{2184}  by 9

Now, we can go two ways. We can either try simplifying this further or we find the cyclicity of Remainder of power of 2 by 9.

Let's learn both the ways

Method 1  : Simplification

2^{2184}  =  2^{3*728}  =  (2^3)^{728}  =  8^{728}  =  (9-1)^{728}

Now, when we expand this expression then all the terms except the last term will be a multiple of 9.
=> All terms except the last term will give 0 as remainder then divided by 9
=> Problem is reduced to what is the remainder when the last term is divided by 9
=> What is the remainder when  2184C2184 * 9^0 * (-1)^ { 2184}  is divided by 9 = Remainder of  (-1)^{2184}  by 9
 => Remainder of 1 by 9 = 1

Method 2  : Cyclicity of Remainder of power of 2 by 9

Remainder of  2^1  (=2) by 9 = 2
Remainder of  2^2  (=4) by 9 = 4
Remainder of  2^3  (=8) by 9 = 8
Remainder of  2^4  (=16) by 9 = 7
Remainder of  2^5  (=32) by 9 = 5
Remainder of  2^6  (=64) by 9 = 1
Remainder of  2^7  (=128) by 9 = 2
Remainder of  2^8  (=256) by 9 = 4

=> Cycle is 6 
So, we will divided the power (2184) by 6 and check the remainder, which is 0
=>   2^{2184}  will give the same remainder by 9 as 2^{Cycle} which is 2^6 and the remainder will be 1

So,  Answer will be A  
Hope it Helps!

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

https://www.youtube.com/watch?v=h1HTEhmUb1w

Fisayofalana · Answer

Bunuel , does this logic work?

The units digit for 11 will always be 1. Raising 1 to any power will always be 1. So can I just divide 1 by 9? I get 1 as the remainder.

chetan2u · Answer

No, that logic is not correct.
Say, the number is 81^11. Thr units digit is 1 but 81 itself is divisible by 9.
Thus, the remainder would depend on entire number and not just the units digit.
It's only when you divide by 10, the remainder would depend on unit's digit.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=87464 --- Harsha GMAT-Club-Forum-p5ku2khu.png

ShubhadeepB · Answer

((11^{12})^{13})^{14} /9 can be rewritten as  (11^{(12)(13)(14)}) /9 , or  (2^{(12)(13)(14)}) /9. If the dividend is greater than the divisor, then we can substitute the remainder of   \frac{11}{9}   in place of 11, i.e., the dividend.

The cyclicity of 2 with respect to 9 needs to be figured out.

Remainder of  (2/9)  = 2
Remainder of  (2^{2}) /9 = 4
Remainder of  (2^{3}) /9 = 8, or ( −1). The concept of negative remainders comes in extremely handy in numerous remainder questions.  
 
Is there a way we can manipulate the powers so that  ((−1)^{even})  appears? Let's check!

(2^{(12)(13)(14)}) /9 =  ((2^{3})^{(4)(13)(14)}) /9 =  (8^{(4)(13)(14)}) /9

This finally simplifies to   ((−1)^{(4)(13)(14)})  /9, or   ((−1)^{even}) /9, which equates to  1/9.

So, the required answer is 1, i.e., choice A.

Regards,
Shubhadeep Basak
GMAT coach
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What is the remainder when ((11^12)^13)^14 is divided by9?

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