Probability with 5 and 7

First look at the cases where $m \gt n$ . (Note that if $m = n$ then we would have $7^{m} + 7^{m} = 2*7^{m}$ , which is clearly not divisible by $5$ as the prime factorization involves only powers of $2$ and $7$ .)

Then $f(m,n) = 7^{n} * (7^{m-n} + 1)$ . This will only be divisible by $5$ if $7^{m-n} \equiv 4\mod{5}$ .

Now look at the sequence of remainders of the powers of $7$ modulus $5$ , starting with power $1$ . We have $2, 4, 3, 1, 2, 4, 3, 1, ....$ . So we have a remainder of $4$ occurring every $4$ th power starting at a power of $2$ . Thus we can have values for $m - n$ of $2, 6, 10, ... , 98$ , a total of $25$ possible values.

Now with both $m, n$ between $1$ and $100$ inclusive we can have $m - n = 2$ in $98$ ways, $m - n = 4$ in $96$ ways, etc., down to having $m - n = 98$ in $2$ ways. This gives us a total of $2 + 4 + 6 + ... + 98 = (\frac{25}{2})(2 + 98) = 1250$ ways in which $m - n$ takes on a value of $2, 6, 10, .... 98$ .

Now the same will hold true for the cases where $n \gt m$ , and thus there are a total of $2*1250 = 2500$ cases where $f(m,n)$ will be divisible by $5$ . As there are $100*100 = 10000$ ordered pairs $(m,n)$ , we end up with a desired probability of

$\dfrac{2500}{10000} = \boxed{0.25}$ .

look at the last digit

7^1=7

7^2=9

7^3=3

7^4=1

the last digit of number that can be divisible by 5, is 5 or 0. Then if the first is 7, the second must be 3 Or if the first is 9, the second must be 1 So, the probability is 4/16=0.25

powers of 7 are 1,2,3,4 (mod 5). So for any 4n consecutive numbers probability remains same.

Now Solve it for 1st 4 natural numbers (The fact that they are 1,2,3,4 (mod 5) helps) to get 0.25

P.S. Real problem would be solving it for 4n+1 or something like first 101 whole numbers