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

$c = a b$

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$b = a c$

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$a = b + c$

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$a = b c$

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Solution

The correct option is D
$a = b c$

$a = 1 + 10 + 10^{2} + . . . . + 10^{54}$

$= \frac{10^{55} - 1}{10 - 1} = \frac{10^{55} - 1}{10^{5} - 1} \times \frac{10^{5} - 1}{10 - 1} = b c$

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a = 111.........1      55, b = 1 + 10 + 10^{2} + 10^{3} + 10^{4}

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