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Q.
If the integers $m$ and $n$ are chosen at random between $1$ and $100$ , then the probability that a number of the form $7^{ m }+7^{ n }$ is divisible by $5$ equals
Probability
Solution:
$7^{ m }+7^{ n }=\left[(5+2)^{ m }+(5+2)^{ n }\right] \equiv 5 \times$ integer $+2^{ m }+2^{ n }$
$\therefore \,\,7^{ m }+7^{ n }$ is divisible by 5 iff $2^{ m }+2^{ n }$ is divisible by $5$ and
so unit place of $2^{ m }+2^{ n }$ must be 0 since it cannot be $5$ . $m $ possible $n$
$1 \,\,\,\,\,3,7, 11,15, .... = 25$
$2\,\,\,\,\, 4, 8,12,..... = 25$
$3 \,\,\,\,\,1, 5, 9, ..... = 25$
$4\,\,\,\,\, 2, 6,10, .... = 25$
Since $2^{1}+2^{3} \equiv 2^{3}+2^{1}$ so $(1,3)$ and $(3,1)$ are same as favourable cases.
$\therefore \,\,\,$required probability$=\frac{25 \times 50}{100 \times 100}=\frac{1}{8}$