KCET · Maths · Binomial Theorem
The digit in the unit's place of \(7^{171}+(177) !\) is
- A 3
- B 2
- C 1
- D 0
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
Given expression
\[
(7)^{171}+(177) !
\]
We know that the cycle of 7 is 4 , i.e., \((7)^{3}=1\) and in case of more the 4 number in factorial notation gives always 0 at unit place.
\[
\Rightarrow \quad\left(7^{3}\right)^{57}+(177) !=1+0=1
\]
\[
(7)^{171}+(177) !
\]
We know that the cycle of 7 is 4 , i.e., \((7)^{3}=1\) and in case of more the 4 number in factorial notation gives always 0 at unit place.
\[
\Rightarrow \quad\left(7^{3}\right)^{57}+(177) !=1+0=1
\]
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