Q40 / 90 · KVPY 2017 (19 Nov SB/SX) · Mathematics
Let $\mathrm{m}, \mathrm{n}$ be two distinct integers chosen randomly from the set $\{0,1,2, \ldots ., 99\}$. Then the probability that $4^{\mathrm{m}}+4^{\mathrm{n}}+3$ is divisible by 5 lies in the interval
- Correct answerA$(0,0.25]$
- B$(0.25,0.5]$
- C$(0.5,0.75]$
- D$(0.75,1)$
Solution
possible case
$\text{Case-1}$
If unit digit is sum of $4^{\mathrm{m}}+4^{\mathrm{n}}$ is 7
possible case are
$\begin{array}{l}(\mathrm{m}, \mathrm{n})=(0,2)(0,4) \ldots \ldots \ldots(0,98)=49 \\ (\mathrm{n}, \mathrm{m})=(0,2)(0,4) \quad \ldots \ldots \ldots . .(0,98)=49 \\ \text { total }=98 \\ \text { Case-2 } \\ \text { Unit digit is sum of } 4^{\mathrm{m}}+4^{\mathrm{n}} \text { is } 2 \\ \text { Possible case } \\ (\mathrm{m}, \mathrm{n})=(2,4)(2,6) \ldots \ldots .(2,98)=48 \\ (\mathrm{~m}, \mathrm{n})=(4,6)(4,8) \ldots \ldots . .(4.98)=47 \\ \vdots \\ (\mathrm{m}, \mathrm{n})=(96,98) \Rightarrow \mathbf{1 C a s e} \\ \text { same case repetition }(\mathrm{n}, \mathrm{m}) \\ 2 \times\left(\frac{48 \times 49}{2}\right) & \\ \text { Now, Case } 1+\text { Case } 2 \\ \Rightarrow 2352+98=2450 \\ \text { Total number of ways } \mathrm{m}, \mathrm{n} \text { can be selected } \\ =100 \times 99 \\ = & 9900 \\ \text { Probability }=\frac{2450}{9900}=0.2474\end{array}$
Hence, the correct answer is option: 'A'.