The remainder when 2740 is divided by 12 is

Question

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

Tardigrade Admin · Accepted Answer

2740= 3120 3119 = (4-1)119 = 119C04119 - 119C1 4118 + 119C24117 - 119C34116+ ...+ (- 1) ∴ 3119 = 4k -1 ∴ 3120 = 12k - 3 = 12 (k- 1) + 9 the required remainder is 9.

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

$2 7^{40} = 3^{120}$
$3^{119} = (4 - 1)^{119} =^{119} C_{0} 4^{119} -^{119} C_{1} 4^{118}$
$+^{119} C_{2} 4^{117} -^{119} C_{3} 4^{116} + ... + (- 1)$
$∴ 3^{119} = 4 k - 1$
$∴ 3^{120} = 12 k - 3 = 12 (k - 1) + 9$
the required remainder is $9$ .

Q. The remainder when 2740 is divided by 12 is

2740=3120 3119=(4−1)119=119C0​4119−119C1​4118 +119C2​4117−119C3​4116+...+(−1) ∴3119=4k−1 ∴3120=12k−3=12(k−1)+9 the required remainder is 9.

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

$2 7^{40} = 3^{120}$
$3^{119} = (4 - 1)^{119} =^{119} C_{0} 4^{119} -^{119} C_{1} 4^{118}$
$+^{119} C_{2} 4^{117} -^{119} C_{3} 4^{116} + ... + (- 1)$
$∴ 3^{119} = 4 k - 1$
$∴ 3^{120} = 12 k - 3 = 12 (k - 1) + 9$
the required remainder is $9$ .