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soumya_joy
- Newbie | Next Rank: 10 Posts
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- Joined: Fri Nov 23, 2012 1:16 pm
(19^26 + 33^26) = (26 - 7)^26 + (26 + 7)^26soumya_joy wrote:The remainder, when (19^26+ 33^26) is divided by 26, is
A. 7 B. 0 C. 15 D. 12
Now, every term except the last term of the expansion of (26 - 7)^26 and (26 + 7)^26 will be a multiple of 26. The last term of both the expansions are 7^26 which is not multiple of 26.
Hence, the remainder will be equal to the remainder when (7^26 + 7^26) = 2*(7^26) is divided by 26.
2*(7^26) = 2*(49^13) = 2*49*(49^12) = 2*49*(52 - 3)^12
Now, every term except the last term of the expansion of (52 - 3)^12 will be a multiple of 52 and hence multiple of 26. The last term of the expansion is 3^12 which is not multiple of 26.
Hence, the remainder will be equal to the remainder when 2*49*(3^12) = 98*(3^12) is divided by 26.
98*(3^12) = 98*(27^4) = 98*(26 + 1)^4
Now, every term except the last term of the expansion of (26 + 1)^4 will be a multiple of 26. The last term of the expansion is 1^4 = 1 which is not multiple of 26.
Hence, the remainder will be equal to the remainder when 98*1 = 98 is divided by 26.
Hence, the remainder is (98 - 3*26) = (98 - 78) = 20
