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Binomial Theorem

Section: Mathematics

Question : 10 of 33

Marks: +1, -0

k

10^k

9^{11}+11^9

k

Solution:

9^{11}+11^9=\;(10-1)^{11}+(10+1)^9

={}^{11}C_0 10^{11}\;+{}^{11}C_1 10^9+\dots {}^{11}C_{11}

\;+{}^{9}C_0 10^{10}+{}^{9}C_1 10^8+\dots+{}^{9}C_9

Sum of last two terms of both expansion as other term have at least

10^2

as a factor

\;(110-1)+90+1=200

\; \therefore \; 9^{11}+11^9=\; \text{Multiple of}\;

\;100+200=\; \text{Multiple of}\;100

Since

200=2 \times 10^2

So,

9^{11}+11^9

is divisible by

10^2

\therefore \; k=2

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