The remainder obtained when 2740 is divided by 12 is

Question

The remainder obtained when 2740 is divided by 12 is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(27)40=(33)40=3120 3119=(4 - 1)119 = 119C04119- 119C14118+ 119C24117-. ldots ldots .+ 119C1184- 119C119 =4k-1 ⇒ 3120=3(4 k - 1)=12k-3 =12(k - 1)+9 So, the required remainder is 9 .

Q. The remainder obtained when $2 7^{40}$ is divided by $12$ is

$(27)^{40} = (3^{3})^{40} = 3^{120}$
$3^{119} = (4 - 1)^{119}$
$=_^{119} C_{0} 4^{119} -_^{119} C_{1} 4^{118} +_^{119} C_{2} 4^{117} - . \dots\dots . +_^{119} C_{118} 4 -_^{119} C_{119}$
$= 4 k - 1$
$\Rightarrow 3^{120} = 3 (4 k - 1) = 12 k - 3$
$= 12 (k - 1) + 9$
So, the required remainder is $9$ .

Q. The remainder obtained when 2740 is divided by 12 is

(27)40=(33)40=3120 3119=(4−1)119 =_119C0​4119−_119C1​4118+_119C2​4117−.…….+_119C118​4−_119C119​ =4k−1 ⇒3120=3(4k−1)=12k−3 =12(k−1)+9 So, the required remainder is 9 .

Q. The remainder obtained when $2 7^{40}$ is divided by $12$ is