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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Basics of Mathematics

NDA 2023 · Mathematics · Basics of Mathematics

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

If 1!+3!+5!+7!+…+1991!+3!+5!+7!+\ldots+199 ! is divided by 24 , what is the remainder?

Options

  1. A3
  2. B6
  3. C7
  4. D9

Correct answer

C. 7

Step-by-step solution

 Since, 1!+3!+5!+7!+…………+199!24=1+6+5×24+7×6×5×24+…………+199 ! 24=1+624+24k24=724+k( Where k∈Z)=7\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}
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