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COMEDK · Maths · 6. Mathematical Induction

The remainder obtained when \((1 !)^{2}+(2 !)^{2}+(3 !)^{2}+\ldots+(100 !)^{2}\) is divided by \(10^{2}\) is

  1. A 14
  2. B 17
  3. C 28
  4. D 27
Verified Solution

Answer & Solution

Correct Answer

(B) 17

Step-by-step Solution

Detailed explanation

Here, terms greater than \(5 !\), i.e. \((5 !)^{2},(6 !)^{2}, \ldots \ldots,(100 !)^{2}\) is divisible by 100 . \(\therefore\) For terms \((5 !)^{2},(6 !)^{2}, \ldots(100 !)^{2}\) remainder is 0 . Now, consider \((1 !)^{2}+(2 !)^{2}+(3 !)^{2}+(4 !)^{2}\) \(=1+4+36+576\) \(=617\)…