What is the remainder when 32^32^32 is divided by 9?

Question

What is the remainder when  32^{32^{32}}  is divided by 9?

A. 7
B. 4
C. 3
D. 2
E. 1

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Bunuel · Answer

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Sumi1010 · Answer

We can write 32^32^32 = (27+5)^32^32 
The above number can be expanded using Binomial Theorem (not in the scope of GMAT test);the expansion has summation in which all the terms have the powers on 27 and 5, but the ONLY last term that has zeroth power of 27 (hence 1) along with highest power of 5 i.e. 5^32^32. Thus all the terms except the last one  5^32^32  are divisible by 9. We need to worry on the remainder of 5^32^32 when divided by 9.

32^32 can be written as (2^5)^32 = 2^160.
We need to know the remainder of 5^32^32

5^1 / 9 : Remainder 5
5^2 / 9 : Remainder 7
5^3 / 9 : Remainder 8
5^4 / 9 : Remainder 4
5^5 / 9 : Remainder 2
5^6 / 9 : Remainder 1

5^7 / 9 : Remainder 5
5^8 / 9 : Remainder 7

Thus, the cyclicity of 5^n when divided by 9 is  6  (i.e.remainder repeats after every 6th power of 5).
NOw, when 32^32 is divided by 6 , we get remainder for 2^32 /6 = 4
Lastly, when 5^4 is divided by 9 , it gives remainder 4.

B

Bunuel 
Do we actually get such problems at GMAT? It seems lengthy one.

ism1994 · Answer

Hi all,

I reasoned as per following (time: 1.5 min), please let me know if it makes sense:

32^32 must have unit number = 6

Now, 32 is elevated to a number whose unit number is 6, let's see what this implies in terms of the unit number of 32^32^32:

if 32^6 ---> unit number must be 4 ---> a number ending with 4 must give reminder 5 when divided by 9 (not included in the options)

if 32^16 ---> unit number must be 6 ---> a number ending with 6 must give reminder 3 when divided by 9

if 32^26 ---> unit number must be 4 ---> a number ending with 4 must give reminder 5 when divided by 9 (not included in the options)

and so on

so IMO ans is C

Thanks

Kinshook · Answer

What is the remainder when  32^{32^{32}}  is divided by 9?

The remainder when  32^{32^{32}}  is divided by 9 
= The remainder when  (-4)^{32^{32}}  is divided by 9 
= The remainder when  4^{32^{32}}  is divided by 9
= The remainder when  16^{16^{32}}  is divided by 9 
= The remainder when  (-2)^{16^{32}}  is divided by 9
= The remainder when  2^{16^{32}}  is divided by 9
= The remainder when  16^{4^{32}}  is divided by 9
= The remainder when  (-2)^{4^{32}}  is divided by 9
= The remainder when  2^{4^{32}}  is divided by 9
= The remainder when  16^{32}  is divided by 9
= The remainder when  (-2)^{32}  is divided by 9
= The remainder when  2^{32}  is divided by 9
= The remainder when  16^8  is divided by 9
= The remainder when  (-2)^8  is divided by 9
= The remainder when  2^8=256  is divided by 9
4

IMO B

KB1920 · Answer

32^32^32= 32^(2^5)^32=32^2^160=32^2^40(4)=> apply cyclicity rule 2^160 will have same unit digit as 2^4=>6=> 32^2^160 will have same unit digit as 32^6 => 4 When 4 divided by 9, remainder=> 4

BrushMyQuant · Answer

We need to find the remainder when  32^{32^{32}}  is divided by 9?

32^{32^{32}} =  32^{2^{5*32}}  =  32^{2^{160}}  
 
 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.

=>  32^{2^{160}}  =  (36-4)^{2^{160}}

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 9
=> All terms except the last term will give remainder of 0 when divided by 9 
=> Problem is reduced to what is the remainder when the last term (i.e. 2^160 C 2^160 * 36^0 *  (-4)^{2^{160}} ) is divided by 9
=> Remainder of  (-4)^{2^{160}}  is divided by 9
=> Remainder of  2^{2^1 * 2^{160}}  is divided by 9
=> Remainder of  2^{2^{161}}  is divided by 9

To solve this problem we need to find the cycle of remainder of power of 2 when divided 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  (=64) by 9 = 2

=> Cycle is 6

We need to find Remainder of  2^{161}  by 6

Remainder of  2^1  (=2) by 6 = 2
Remainder of  2^2  (=4) by 6 = 4
Remainder of  2^3  (=8) by 6 = 2
Remainder of  2^4  (=16) by 6 =  4

=> Cycle is 2

=> Remainder of  2^{161}  by 6 = Remainder of  2^{1}  by 6 = 2
 => Remainder of  2^{2^{161}}  is divided by 9 = Remainder of 2^2 by 9 = 4 
 
So,  Answer will be B 
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

What is the remainder when 32^32^32 is divided by 9?

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What is the remainder when 32^32^32 is divided by 9?

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