The minimum positive integral value of $$m$$ such that $${\left( {1073} \right)^{71}} - m$$ may be divisible by 10, is
A.
1
B.
3
C.
7
D.
9
Answer :
7
Solution :
$$\eqalign{
& {\left( {1073} \right)^{71}} - m = {\left( {73 + 1000} \right)^{71}} - m \cr
& = {\,^{71}}{C_0}{\left( {73} \right)^{71}} + {\,^{71}}{C_1}{\left( {73} \right)^{70}}\left( {1000} \right) + {\,^{71}}{C_2}{\left( {73} \right)^{69}}{\left( {1000} \right)^2} + ..... + {\,^{71}}{C_{71}}{\left( {1000} \right)^{71}} - m \cr} $$
Above will be divisible by 10 if $$^{71}{C_0}{\left( {73} \right)^{71}}$$ is divisible by 10
Now, $$^{71}{C_0}{\left( {73} \right)^{71}} = {\left( {73} \right)^{70}} \cdot 73 = {\left( {{{73}^2}} \right)^{35}} \cdot 73$$
The last digit of $$73^2$$ is 9, so the last digit of $${\left( {{{73}^2}} \right)^{35}}$$ is 9.
$$\therefore $$ Last digit of $${\left( {{{73}^2}} \right)^{35}} \cdot 73{\text{ is }}7$$
Hence, the minimum positive integral value of $$m$$ is 7, so that it is divisible by 10.
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is