Question

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

Given : a = 111 ... 1 (55 digits), b = 1 + 10+10^2 +......+10^4,c=1+ 10^5 +10^10 +10^15.....+10^50,

To Find : correct option :

a=b+c   .

a=bc

b=ac

c=ab

Solution:

b = 1 + 10 + 10²  + 10³  + 10⁴

c  = 1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+ 10⁵⁰

bc = ( 1 + 10 + 10²  + 10³  + 10⁴ ) (1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+ 10⁵⁰ )

=1 (1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+ 1010⁵⁰ ) + 10 (1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+ 10⁵⁰ )+  10² (1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+10⁵⁰ )+  10³ (1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+ 10⁵⁰ )+  10⁴ (1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+ 10⁵⁰ )

=  (1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+ 10⁵⁰ )  + (10¹ + 10⁶ + 10¹¹ + 10¹⁶ + ...............+ 10⁵¹) +  (10²+ 10⁷ + 10¹² + 10¹⁷ + ...............+ 10⁵²) +  (10³+ 10⁸ + 10¹³ + 10¹⁸ + ...............+ 10⁵³) +  (10⁴+ 10⁹ + 10¹⁴ + 10¹⁹ + ...............+ 10⁵⁴)

= 1 + 10¹  + 10² + ...................................+ 10⁵⁴

= 10⁵⁴ + 10⁵³ + ....................+ 10² +  10¹  + 1

= 111...........1  ( 55 Digits)

Hence a = bc

Another way

b = 1 + 10 + 10²  + 10³  + 10⁴   =  1 (10⁵ - 1)/(10 - 1)  = (10⁵ - 1)/9

c =  1 + 10⁵ + 10¹⁰ + 10¹⁵ + ...............+ 10⁵⁰ = 1 ( (10⁵)¹¹ - 1) /(10⁵ - 1) = (10⁵⁵ - 1)/(10⁵ - 1)

=> bc = (10⁵⁵ - 1)/9

= (10⁵⁵ - 1)/(10 - 1)

= 1111.....1  ( 55 times)   as   1  n times=  (  (10ⁿ - 1)/(10 - 1)

Hence bc = a  

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