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Question

If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the from 7m+7n is divisible by 5, equals

A
14
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B
17
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C
18
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D
149
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Solution

The correct option is A 14
71=7, 72=49, 73=343, 74=2401,
Therefore, for 7r, r ϵ N the number ends with 7, 9, 3, 1, 7, . . .
7m+7n will be divisible by 5 if it ends in 0 or 5.
The last digit of m and n belongs to the set {1,3,7,9}
Hence there are 4×4=16 combinations of digits which determines the last digit of the sum.
The favourable cases among them are { (1,9),(9,1),(3,7),(7,3) }

Hence, the required probability is
P=416=14

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