Latest self Aptitude Question SOLUTION: find the remainder of 32^32^32 when divided by 9

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find the remainder of 32^32^32 when divided by 9

niraj kumar (11 years ago)

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sumit kumar  (11 years ago)

32 = 2⁵
(2⁵)³² = 2¹⁶⁰
(2¹⁶⁰)³² = 2⁵¹²⁰
You are looking for the remainder of 2⁵¹²⁰ ÷ 9.

2¹ mod 9 = 2
2² mod 9 = 4
2³ mod 9 = 8
2⁴ mod 9 = 7
2⁵ mod 9 = 5
2⁶ mod 9 = 1
2⁷ mod 9 = 2
2⁸ mod 9 = 4
2⁹ mod 9 = 8
2¹⁰ mod 9 = 7
2¹¹ mod 9 = 5
2¹² mod 9 = 1
...
Note that the pattern repeats, and when the exponent is divisible by 6, the remainder is 1.
5120 = 6(853) + 2, therefore
2⁵¹²⁰ mod 9 = 4

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K Vedant Laskar  (11 years ago)

The answer is 7

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DIVYA SHARMA  (9 years ago)

K VEDANT LASKAR
PLEASE POST THE SOLUTION

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Amrendra kumar  (9 years ago)

32^32^32/9 can be written as
5^32^32
25^16^32
7^16^32
49^8^32
4^8^32
16^4^32
7^4^32
49^2^32
4^2^32
16^32
7^32
49^16
4^16
16^8
7^8
49^4
4^4
16^2
7^2
49when divided by 9gives 4 as a ans

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