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+7ⁿ is divisible by 5 is?

Answer is given as 1/5

Plz explain

Dear student,

If a number is divisible by 5 - it should either end at 0 or 5

the possible values last digit of 7^m is = 7,9,3,1

one can select two numbers from it in = 6 ways

2 favourable conditions are = 7,3 and 9,1

Probability =1/3

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Askiitians Expert
Sagar Singh
B.Tech, IIT Delhi

The answer is 1/5. The total possibilities are....(7,3), (7,9), (7,1), (3,9),(3,1),(9,1),(7,7),(9,9),(3,3),(1,1).......

Among these oly (7,3) and (9,1) satisfy the condition.....so we can take the probability as 2/10 ie 1/5

I guess it will be 1/4 .Cases are (7,7)(7,9)(9,7)(7,3)(3,7)..............(1,1)total cases are 16 and favourable cases are 4 . so it`s 1/4