• # question_answer 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 A) $\frac{1}{4}$ B) $\frac{1}{7}$ C) $\frac{1}{8}$   D) $\frac{1}{49}$

Solution :

 ${{7}^{m}}+{{7}^{n}}=[{{(5+2)}^{m}}+{{(5+2)}^{n}}]$$\equiv 5\times \operatorname{int}egar+{{2}^{m}}+{{2}^{n}}$ $\therefore$${{7}^{m}}+{{7}^{n}}$is divisible by 5  i ff ${{2}^{m}}+{{2}^{n}}$ 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 $=\frac{25\times 50}{100\times 100}=\frac{1}{8}$

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