Let m, n be two distinct integers chosen rand | vedclass

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(A) The total number of ways to choose two distinct integers $m$ and $n$ from the set $\{0, 1, 2, \ldots, 99\}$ is given by $\binom{100}{2} = \frac{10

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$(0, 0.25]$

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(A) The total number of ways to choose two distinct integers $m$ and $n$ from the set $\{0, 1, 2, \ldots, 99\}$ is given by $\binom{100}{2} = \frac{100 	imes 99}{2} = 4950$.We want $4^m + 4^n + 3 \equiv 0 \pmod{5}$.Since $4 \equiv -1 \pmod{5}$,we have $4^m + 4^n + 3 \equiv (-1)^m + (-1)^n + 3 \pmod{5}$.For this expression to be divisible by $5$,we need $(-1)^m + (-1)^n + 3 \equiv 0 \pmod{5}$,which implies $(-1)^m + (-1)^n \equiv -3 \equiv 2 \pmod{5}$.Since $(-1)^m$ and $(-1)^n$ can only be $1$ or $-1$,the only way for their sum to be $2$ is if $(-1)^m = 1$ and $(-1)^n = 1$.This means both $m$ and $n$ must be even integers.In the set $\{0, 1, 2, \ldots, 99\}$,there are $50$ even integers $(	ext{i.e.}, 0, 2, 4, \ldots, 98)$.The number of ways to choose two distinct even integers is $\binom{50}{2} = \frac{50 	imes 49}{2} = 1225$.The probability is $P = \frac{1225}{4950} = \frac{49}{198} \approx 0.24747$.Since $0.24747 \in (0, 0.25]$,the correct interval is $(0, 0.25]$.

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

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