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04 Dec 2020, 07:20
Kudos
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B
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Difficulty:
95%
(hard)
Question Stats:
41% (02:09) correct
59%
(02:11)
wrong
based on 166
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What is the remainder when \(32^{32^{32}}\) is divided by 9?
A. 7
B. 4
C. 3
D. 2
E. 1
Are You Up For the Challenge: 700 Level Questions
A. 7
B. 4
C. 3
D. 2
E. 1
Are You Up For the Challenge: 700 Level Questions
04 Dec 2020, 07:22
Kudos
Bookmarks
04 Dec 2020, 09:09
Kudos
Bookmarks
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.
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.
