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28 Jan 2017, 18:41
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What is the remainder when 19^(23^24) is divided by 16?
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Sent from my Lenovo A6020a46 using GMAT Club Forum mobile app
01 Feb 2017, 17:30
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Harshit401
Hi Harshit401,
When it is divided by 16, the remainder of 19^1 is 3.
When it is divided by 16, the remainder of 19^2 is 9.
When it is divided by 16, the remainder of 19^3 is 11.
When it is divided by 16, the remainder of 19^4 is 1.
Hence the period of (19^n)'s remainder is 4, when it is divided by 16.
When it is divided by 4, the remainder of 23^1 is 3.
When it is divided by 4, the remainder of 23^2 is 1.
Hence the period of (23^n)'s remainder is 2, when it is divided by 4.
The remainder of 23^24 is 1, when it is divided by 4.
The remainder of 19^(23^24) is equal to 19^1, it is 3, when it is divided by 16.
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