Key Concepts and Formulas

• Divisibility rule: A number is divisible by 5 if its last digit is either 0 or 5.
• Counting principle: If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.
• Permutations: Since $x$ and $y$ are distinct, the order matters.

Step-by-Step Solution

Step 1: Categorize the integers based on their remainders when divided by 5.

We divide the integers from 1 to 25 into 5 categories based on their remainders when divided by 5.

• Numbers of the form $5\lambda$: 5, 10, 15, 20, 25 (5 numbers)
• Numbers of the form $5\lambda + 1$: 1, 6, 11, 16, 21 (5 numbers)
• Numbers of the form $5\lambda + 2$: 2, 7, 12, 17, 22 (5 numbers)
• Numbers of the form $5\lambda + 3$: 3, 8, 13, 18, 23 (5 numbers)
• Numbers of the form $5\lambda + 4$: 4, 9, 14, 19, 24 (5 numbers)

Step 2: Determine the pairs of remainders that sum to a multiple of 5.

For $x + y$ to be divisible by 5, the sum of their remainders when divided by 5 must be divisible by 5. The possible pairs of remainders are:

• 0 + 0
• 1 + 4
• 2 + 3
• 3 + 2
• 4 + 1

Step 3: Count the number of ways to choose $x$ and $y$ for each pair of remainders.

• Case 1: Both $x$ and $y$ are of the form $5\lambda$. There are 5 choices for $x$. Since $x$ and $y$ must be distinct, there are 4 choices for $y$. So, there are $5 \times 4 = 20$ ways.

• Case 2: $x$ is of the form $5\lambda + 1$ and $y$ is of the form $5\lambda + 4$. There are 5 choices for $x$ and 5 choices for $y$. So, there are $5 \times 5 = 25$ ways.

• Case 3: $x$ is of the form $5\lambda + 2$ and $y$ is of the form $5\lambda + 3$. There are 5 choices for $x$ and 5 choices for $y$. So, there are $5 \times 5 = 25$ ways.

• Case 4: $x$ is of the form $5\lambda + 3$ and $y$ is of the form $5\lambda + 2$. There are 5 choices for $x$ and 5 choices for $y$. So, there are $5 \times 5 = 25$ ways.

• Case 5: $x$ is of the form $5\lambda + 4$ and $y$ is of the form $5\lambda + 1$. There are 5 choices for $x$ and 5 choices for $y$. So, there are $5 \times 5 = 25$ ways.

Step 4: Calculate the total number of ways.

The total number of ways is $20 + 25 + 25 + 25 + 25 = 120$.

Common Mistakes & Tips

• Remember to account for the condition that $x$ and $y$ are distinct.
• Be careful when counting the number of elements in each category.
• Don't forget to consider all possible pairs of remainders that sum to a multiple of 5.

Summary

We categorized the integers from 1 to 25 based on their remainders when divided by 5. Then we found the pairs of remainders that sum to a multiple of 5. Finally, we counted the number of ways to choose $x$ and $y$ for each pair of remainders, remembering that $x$ and $y$ must be distinct. Summing the counts for each case gives the total number of ways.

The final answer is \boxed{120}.