The remainder when 2740 is divided by 12 is

Question

The remainder when 2740 is divided by 12 is (A) 3 (B) 7 (C) 9 (D) 11

Tardigrade Admin · Accepted Answer

2740=3120 ⇒ 3119=(4-1)119 = 119 C 0 4119- 119 C 1 4118 +119 C 2 4117- 119 C 3 4116+ ldots .+(-1) ∴ 3119=4 k-1 ∴ 3120=12 k-3=12(k-1)+9 ∴ The required remainder is 9

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

3

7

9

11

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

Q. The remainder when 2740 is divided by 12 is

3

7

9

11

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

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