<s>[INST] 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 is 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 is
$A. 1/4$
$B. 1/7$
$C. 1/8$
$D. 1/49$  
I did this: Let $m>n$ (Clearly m and n can't be equal because $5$ can't divide $2*7^{m}$). 
Now $7^{m}+7^{n}=7^{n}(7^{m-n}+1)$. If $5$ has to divide this, it implies that 5 has to divide $(7^{m-n}+1)$ (because it cannot divide a power of $7$). Since powers of $7$ have a cyclic order of $7,9,3,1$; $7^{m-n}$ has to therefore end in a $9$ and therefore $m-n$ can be $2,6,10,...,98$. Hence the only set of values of $n$ are $(1,2,3,...,97),(1,2,3,...,93),(1,2,3,...,89),...,(1)$. Also fixing $n$ would fix $m$.  
Therefore the number of favorable cases is $97+93+89+...+2=1224$. Which means that the required probability should be $1224/100C2$ which turns out to be $68/275$, which is not matching with any of the options..  Where did I go wrong ??
Please help !!
 [/INST] Shouldn't the answer be $24/100$? This is because there are 40 numbers which satisfy $7^x =  2^x = 1 \mod 5$, and $20$ each for $7^x = 2^x = 2,3,4 \mod 5$, so we can't just assign equal probabilities to each event.
( Saying equal likeliness of the events $1+4,2+3,3+2,4+1$ as the sum of the remainders out of $4\cdot 4=16$ possible combinations).
The answer should be 
$$
\frac{40}{100} \cdot \frac{20}{100} + \frac{20}{100} \cdot \frac{20}{100} + \frac{20}{100} \cdot \frac{20}{100} + \frac{20}{100} \cdot \frac{40}{100} = \frac{24}{100}.
$$
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