Question

Let a = 111...1(55 digits),
$b=1+10+{10}^2+{10}^3+{10}^4$
$c=1+{10}^5+{10}^{10}+{10}^{15}+...+{10}^{50},then$

• A. a = b + c
• B. a = bc
• C. b = ac
• D. c = ab

Answer 1

The correct option is B a = bc

Since, a = 1111...1(55 digits)
$b=1+10+{10}^2+{10}^3+{10}^4=\frac{1({10}^5-1)}{10-1}=\frac{{10}^5-1}{9}andc=1+{10}^5+{10}^{10}+{10}^{15}+...{10}^{50}=1\frac{\left[\right({10}^5{)}^{11}-1]}{{10}^5-1}bc=\frac{({10}^5-1)}{9}\times \frac{({10}^{55}-1)}{{10}^5-1}=\frac{\left(\mathrm{9999.....55}digits\right)}{9}=a$