Animated Solution for Mathematics - Probability: If the integers $m$ and $n$ are chosen at random from 1 to 100, then the probability that a number of the form $7^m+7^n$ is divisible by 5 equals

To determine the probability that the sum $7^m+7^n$ is divisible by $5$, where $m,n\in \{1,2,\dots ,100\}$, we must focus on the units digit of the powers of $7$. A number is divisible by $5$ if and only if its units digit is $0$ or $5$.

Let us examine the units digit of $7^k$ for increasing values of $k$: - For $k=1$, $7^1=7$, units digit is $7$. - For $k=2$, $7^2=49$, units digit is $9$. - For $k=3$, $7^3=343$, units digit is $3$. - For $k=4$, $7^4=2401$, units digit is $1$.

The sequence of units digits follows a cycle of length $4$: $\{7,9,3,1\}$. Since $100$ is perfectly divisible by $4$, each of these four digits appears with equal frequency ($25$ times each) as $m$ and $n$ range from $1$ to $100$. Thus, the probability of any power $7^k$ ending in $7,9,3,$ or $1$ is exactly $\frac{1}{4}$.

We must determine which pairs of units digits $(u_m,u_n)$ result in a sum $u_m+u_n$ that ends in $0$ or $5$. Note that every power of $7$ is an odd number.

The sum of two odd numbers, $7^m+7^n$, is always an even number. An even number can never end in $5$. Therefore, the sum $7^m+7^n$ is divisible by $5$ if and only if its units digit is $0$.

We now identify the pairs $(u_m,u_n)$ from the set $\{7,9,3,1\}$ such that $u_m+u_n$ ends in $0$: - $7+3=10$ - $3+7=10$ - $9+1=10$ - $1+9=10$

There are exactly $4$ valid pairs out of the $4\times 4=16$ possible combinations of units digits. Since the choices of $m$ and $n$ are independent and each digit in the cycle is equally likely, the probability is calculated as:

The final probability that $7^m+7^n$ is divisible by $5$ is $\frac{1}{4}$.