If the integers m and n r chosen at random with replacement frm 1 to 100 , then what is the probability that a no. of the form 7^m + 7^n is divisible by 5?
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1 Answers
49on close thinking we can see than
74k ends with 1
74k+1 ends with 7
74k+2 ends with 9
74k+3 ends with 3
where k belongs to {0,1,2,3,4,5,6,7,8,9,10......∞}
thus, for 7m+7n to be divisible by 5,
m must be of form 4k while n is of form 4k+2 or vice versa.
Or m must be of form 4k+1 while n is of form 4k+3 or vice versa.
so P(E)=2x25x25+2x25x2550x99=5099
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