If $p$ is selected from the set ${1, 2, 3, . . . . . .100}$ , then the probability of selecting $p$ such that the number $2^{p} + 3^{p} + 5^{p}$ is divisible by $4$ is

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Solution

For $p = 1$
$2 + 3 + 5 = 10$ which is not divisible by $4$ .
For $p \geq 2$
$2^{p} + (4 - 1)^{p} + (4 + 1)^{p} = (- 1)^{p} + (1)^{p} + 4 n; n \in I$
So this will be divisible when $p$ is odd except $1$
Required number of $p$ is $49$
Total possible values of $p$ is $100$
$P (e) = \frac{49}{100} = 0.49$

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If p is selected from the set {1,2,3,......100}, then the probability of selecting p such that the number 2p+3p+5p is divisible by 4 is

For p=1 2+3+5=10 which is not divisible by 4. For p≥2 2p+(4−1)p+(4+1)p=(−1)p+(1)p+4n ; n∈I So this will be divisible when p is odd except 1 Required number of p is 49 Total possible values of p is100 P(e)=49100=0.49

If $p$ is selected from the set ${1, 2, 3, . . . . . .100}$ , then the probability of selecting $p$ such that the number $2^{p} + 3^{p} + 5^{p}$ is divisible by $4$ is