COMEDK · Maths · 6. Mathematical Induction
The last digit of \(583 !+7^{291}\) is
- A 1
- B 2
- C 0
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
We have, \(583 ! \equiv 0(\bmod 10)\) and \(\quad 7^{2}=-1(\bmod 10)\) So, \(\left(7^{2}\right)^{145} \cdot 7^{1}=(-1)^{145} \cdot 7^{1}(\bmod 10)\) \(\Rightarrow \quad 7^{291}=-7(\bmod 10) \equiv 3(\bmod 10)\) So, \(583 !+7^{291}=3(\bmod 10)\) Hence, last digit is \(3 .\)
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