Q152 / 270 · NDA 2023 (Phase 2) · Mathematics

If $1!+3!+5!+7!+\ldots+199$ ! is divided by 24 , what is the remainder?

  • A
    3
  • B
    6
  • Correct answer
    C
    7
  • D
    9

Solution

$\begin{aligned} & \text { Since, } \frac{1!+3!+5!+7!+\ldots \ldots \ldots \ldots+199!}{24} \\ & =\frac{1+6+5 \times 24+7 \times 6 \times 5 \times 24+\ldots \ldots \ldots \ldots+199 \text { ! }}{24} \\ & =\frac{1+6}{24}+\frac{24 \mathrm{k}}{24}=\frac{7}{24}+\mathrm{k}(\text { Where } \mathrm{k} \in \mathrm{Z})=7 \end{aligned}$

Hence, the correct answer is option: 'C'.

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