A) 7
B) 9
C) 1
D) 3
Correct Answer: C
Solution :
We have \[{{7}^{2}}=49=50-1\] Now, \[{{7}^{300}}={{({{7}^{2}})}^{150}}={{(50-1)}^{150}}\] = \[^{150}{{C}_{0}}{{(50)}^{150}}{{(-1)}^{0}}+{{\,}^{150}}{{C}_{1}}{{(50)}^{149}}{{(-1)}^{1}}+......\]\[+{{\,}^{150}}{{C}_{150}}{{(50)}^{0}}{{(-1)}^{150}}\] Thus the last digits of \[{{7}^{300}}\] are \[^{150}{{C}_{150}}.1.1\] i.e., 1.
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