Let a = 1 1 1 ... 1 (55 digits), b = 1 + 10 + 102 + ... + 104, c = 1 + 105 + 1010 + 1015 + ... + 1050, then

a=1+10+102+⋯+1054 =1055−110−1=1055−1105−1×105−110−1=bc

Let a = 1 1 1 ... 1 (55 digits), b = 1 + 10 + 102 + ... + 104, c = 1 + 105 + 1010 + 1015 + ... + 1050, then

a=1+10+102+⋯+1054 =1055−110−1=1055−1105−1×105−110−1=bc

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Let a = 1 1 1 ... 1 (55 digits), b = 1 + 10 + 10² + ... + 10⁴, c = 1 + 10⁵ + 10¹⁰ + 10¹⁵ + ... + 10⁵⁰, then

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a = b + c

a = bc

b = ac

c = ab

answer is B.

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$\begin{matrix} a = 1 + 10 + 10^{2} + \dots + 10^{54} \\ = \frac{10^{55} - 1}{10 - 1} = \frac{10^{55} - 1}{10^{5} - 1} \times \frac{10^{5} - 1}{10 - 1} = bc \end{matrix}$

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Let a = 1 1 1 ... 1 (55 digits), b = 1 + 10 + 102 + ... + 104, c = 1 + 105 + 1010 + 1015 + ... + 1050, then

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Let a = 1 1 1 ... 1 (55 digits), b = 1 + 10 + 10² + ... + 10⁴, c = 1 + 10⁵ + 10¹⁰ + 10¹⁵ + ... + 10⁵⁰, then