MCQ
Let $m, n$ be two distinct integers chosen randomly from the set $\{0,1,2, \ldots, 99\}$. Then, the probability that $4^m+4^n+3$ is divisible by $5$ lies in the interval
- ✓$(0,0.25]$
- B$(0.25,0.5]$
- C$(0.5,0.75]$
- D$(0.75,1)$
We have,
$m, n$ be two distinct integer chosen
randomly from the set $\{0,1,2,3, \ldots, 99\}$
$\therefore$ Total number of outcomes $=100\,C_2$
$4^m+4^n+3$
$(5-1)^m+(5-1)^n+3$
$5 K+(-1)^m+5 \lambda+(-1)^n+3$
$5(K+\lambda)+(-1)^m+(-1)^n+3$
$4^m+4^n+3$ is divisible by 5 if $m$ and $n$ both are even.
$\therefore$ Favourable outcomes $=50\,C_2$
Required probability $=\frac{50\,C_2}{100\,C_2}$
$=\frac{50 \times 49}{100 \times 99}=\frac{49}{198}=0.247$
$0.247 \in(0,0.25]$
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