COMEDK · Maths · 6. Mathematical Induction
The remainder obtained when \((1 !)^{2}+(2 !)^{2}+(3 !)^{2}+\ldots+(100 !)^{2}\) is divided by \(10^{2}\) is
- A 14
- B 17
- C 28
- D 27
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 \((1 !)^{2}+(2 !)^{2}+(3 !)^{2}+(4 !)^{2}\) \(=1+4+36+576\) \(=617\)…
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