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4412 ÷ 7 = (42 + 2)12
By using binomial theorem, we have
(42 + 2)12 = 12C0 × 4212 × 20 + 12C1 × 4211 × 21 + …………+ 12C12 × 420 × 212
So, in the above expression only the last term will not be completely divisible by ‘7’.
So, required remainder = Remainder obtained when 12C12 × 420 × 212 is divided by ‘7’ or 212 is divided by ‘7’.
Now, 212 ÷ 7 = (23)4 ÷ 7 = 84 ÷ 7 = (7 + 1)4 ÷ 7
So, again by binomial expansion, we get
(7 + 1)4 = 4C0 × 712 × 10 + 4C1 × 711 × 11 + …………+ 4C4 × 70 × 14
So, last term will not be divisible by ‘7’.
So, required remainder = 4C4 × 70 × 14 = 1
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