KVPY Sample Paper KVPY Stream-SX Model Paper-25

  • 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) \[\frac{1}{49}\]

    Correct Answer: C

    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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