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COMEDK · Maths · 6. Mathematical Induction

The last digit of \(583 !+7^{291}\) is

  1. A 1
  2. B 2
  3. C 0
  4. D 3
Verified Solution

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 \(3 .\)