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The digit at units place in the number 17^{1995}+11^{1995}-7^{1995} is 

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

17^{1995}+11^{1995}-7^{1995}=(7+10)^{1995}+(1+10)^{1995}-7^{1995}

\begin{aligned} =\left[7^{1995}+{ }^{1995} \mathrm{C}_1 7^{1994} \cdot 10^1+{ }^{1995} \mathrm{C}_2 7^{1993} \cdot 10^2+\ldots .+{ }^{1995} \mathrm{C}_{1995} \cdot{ }^{} 10^{1995}\right] \\ +\left[{ }^{1995} \mathrm{C}_0{}+{ }^{1995} \mathrm{C}_1 10^1+{ }^{1995} \mathrm{C}_2 10^2+\ldots \ldots+{ }^{1995} \mathrm{C}_{1995} 10^{1995}\right]-7^{1995} \end{aligned}

\begin{aligned} & =\left[{ }^{1995} \mathrm{C}_1 7^{1994} \cdot 10^1+\ldots .+10^{1995}\right]+\left[{ }^{1995} \mathrm{C}_1 10^1+\ldots . .+{ }^{1995} \mathrm{C}_{1995} 10^{1995}\right]+{ }^{1995} \mathrm{C}_0 \\ & =(\text { a multiple of } 10)+1 \end{aligned}

Thus, the units place digits is 1.

Posted by

shivangi.bhatnagar

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