If (1/2 ⋅ 310)+(1/22 ⋅ 39)+ ldots (1/210 ⋅ 3)=( K /210 ⋅ 310), then the remainder when K is divided by 6 is

Question

If (1/2 ⋅ 310)+(1/22 ⋅ 39)+ ldots (1/210 ⋅ 3)=( K /210 ⋅ 310), then the remainder when K is divided by 6 is (A) 1 (B) 2 (C) 3 (D) 5

Tardigrade Admin · Accepted Answer

(1/2 ⋅ 310)+(1/22 ⋅ 39)+(1/23 ⋅ 38)+ ldots .+(1/210 ⋅ 3)=(K/210 ⋅ 310) K =29+28 ⋅ 3+27 ⋅ 32+ ldots ldots+39 =(29(((3/2))10-1)/(3)2-1)=310-210 Now, 310-210=(35-25)(35+25) =(211)(275) =(35 × 6+1)(45 × 6+5) =6 λ+5 Remainder is 5 .

Q. If 2⋅3101​+22⋅391​+…210⋅31​=210⋅310K​, then the remainder when K is divided by 6 is

1

2

3

5

2⋅3101​+22⋅391​+23⋅381​+….+210⋅31​=210⋅310K​ K=29+28⋅3+27⋅32+……+39 =23​−129((23​)10−1)​=310−210 Now, 310−210=(35−25)(35+25) =(211)(275) =(35×6+1)(45×6+5) =6λ+5 Remainder is 5.

Q. If $\frac{1}{2 \cdot 3 ^{10}} + \frac{1}{2 ^{2} \cdot 3 ^{9}} + \dots \frac{1}{2 ^{10} \cdot 3} = \frac{K}{2 ^{10} \cdot 3 ^{10}}$ , then the remainder when $K$ is divided by $6$ is