KVPY #8

Algebra Level 5

If 1 0 9 + 2 ( 1 1 1 ) ( 1 0 8 ) + 3 ( 1 1 2 ) ( 1 0 7 ) + ⋯ + 10 ( 1 1 9 ) = 1 0 9 ⋅ k 10^{9} + 2(11^1)(10^8) + 3(11^2)(10^7) + \cdots+ 10(11^9) = 10^9\cdot k , then find the value of k k .

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The answer is 100.

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

Chew-Seong Cheong
Mar 27, 2017

S = 1 0 9 + 2 ( 1 1 1 ) ( 1 0 8 ) + 3 ( 1 1 2 ) ( 1 0 7 ) + . . . + 9 ( 1 0 1 ) ( 1 1 8 ) + 10 ( 1 1 9 ) 11 S 10 = ( 1 1 1 ) ( 1 0 8 ) + 2 ( 1 1 2 ) ( 1 0 7 ) + 3 ( 1 1 3 ) ( 1 0 6 ) + . . . + 9 ( 1 1 9 ) + 1 1 10 S − 11 S 10 = 1 0 9 + ( 1 1 1 ) ( 1 0 8 ) + ( 1 1 2 ) ( 1 0 7 ) + . . . + 1 1 9 − 1 1 10 11 10 ( S − 11 S 10 ) = ( 1 1 1 ) ( 1 0 8 ) + ( 1 1 2 ) ( 1 0 7 ) + ( 1 1 3 ) ( 1 0 6 ) + . . . + 1 1 10 10 − 1 1 11 10 ( S − 11 S 10 ) ( 1 − 11 10 ) = 1 0 9 − 1 1 10 − 1 1 10 10 + 1 1 11 10 S 100 = 1 0 9 − 1 1 10 + 1 1 10 ( 11 − 1 ) 10 = 1 0 9 ⟹ S = 100 ( 1 0 9 ) \begin{aligned} S & = 10^9 + 2(11^1)(10^8) + 3(11^2)(10^7)+...+9(10^1)(11^8)+10(11^9) \\ \frac {11S}{10} & = (11^1)(10^8) + 2(11^2)(10^7) + 3(11^3)(10^6)+...+9(11^9)+11^{10} \\ S - \frac {11S}{10} & = 10^9 + (11^1)(10^8) + (11^2)(10^7)+...+11^9-11^{10} \\ \frac {11}{10}\left(S - \frac {11S}{10}\right) & = (11^1)(10^8) + (11^2)(10^7) + (11^3)(10^6)+...+\frac {11^{10}}{10}-\frac {11^{11}}{10} \\ \left(S - \frac {11S}{10}\right) \left(1-\frac {11}{10}\right) & = 10^9 - 11^{10} - \frac {11^{10}}{10} + \frac {11^{11}}{10}\\ \frac S{100} & = 10^9 - 11^{10} + \frac {11^{10}(11-1)}{10} \\ & = 10^9 \\ \implies S & = 100(10^9) \end{aligned}

⟹ k = 100 \implies k = \boxed{100}

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Thank you sir for uploading such a nice solution...

Rahil Sehgal - 4 years, 2 months ago

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