Question
($7^19$ + 2) is divided by 6, the remainder is :
Answer: Option A
Answer: (a)By the Binomial expansion we have $(x + 1)^n$ =$x^n$ +$n_c_1 x^(n–1)$+$n_c_2 x^(n– 2)$ + ..... + $nc_{n-1}$ x +1 Here, each term except the last term contains x. Obviously, each term except the last term is exactly divisible by x. Following the same logic, $7^19$ = $(6 + 1)^19$ has each term except last term divisible by 6. Hence,$ 7^19$ + 2 when divided by 6 leaves remainder =1 + 2 = 3
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Answer: (a)By the Binomial expansion we have $(x + 1)^n$ =$x^n$ +$n_c_1 x^(n–1)$+$n_c_2 x^(n– 2)$ + ..... + $nc_{n-1}$ x +1 Here, each term except the last term contains x. Obviously, each term except the last term is exactly divisible by x. Following the same logic, $7^19$ = $(6 + 1)^19$ has each term except last term divisible by 6. Hence,$ 7^19$ + 2 when divided by 6 leaves remainder =1 + 2 = 3
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