Let x and y be distinct integers where `1le x le25` and `1 =< y le 25`. Then, the number of ways of choosing x and y, such that x + y is divisible by 5, is _______ .

Answer

To solve the problem, we need to find the number of ways to choose two distinct integers \( x \) and \( y \) from the set of integers from 1 to 25 such that the sum \( x + y \) is divisible by 5.

### Step 1: Classify integers based on their remainder when divided by 5
The integers from 1 to 25 can be classified based on their remainders when divided by 5. The possible remainders are 0, 1, 2, 3, and 4.

- Remainder 0: {5, 10, 15, 20, 25} (5 numbers)
- Remainder 1: {1, 6, 11, 16, 21} (5 numbers)
- Remainder 2: {2, 7, 12, 17, 22} (5 numbers)
- Remainder 3: {3, 8, 13, 18, 23} (5 numbers)
- Remainder 4: {4, 9, 14, 19, 24} (5 numbers)

### Step 2: Identify pairs that sum to a multiple of 5
For \( x + y \) to be divisible by 5, the pairs of remainders must satisfy the following conditions:
- (0, 0)
- (1, 4)
- (2, 3)

### Step 3: Count the combinations for each case
1. **Case (0, 0)**:
- We can choose 2 from the 5 numbers with remainder 0.
- The number of ways to choose 2 from 5 is given by \( \binom{5}{2} = 10 \).

2. **Case (1, 4)**:
- We can choose 1 from the 5 numbers with remainder 1 and 1 from the 5 numbers with remainder 4.
- The number of ways is \( 5 \times 5 = 25 \).

3. **Case (2, 3)**:
- Similarly, we can choose 1 from the 5 numbers with remainder 2 and 1 from the 5 numbers with remainder 3.
- The number of ways is \( 5 \times 5 = 25 \).

### Step 4: Total the combinations
Now, we sum the number of ways from all cases:
- From case (0, 0): 10 ways
- From case (1, 4): 25 ways
- From case (2, 3): 25 ways

Total = \( 10 + 25 + 25 = 60 \).

### Final Answer
The total number of ways to choose \( x \) and \( y \) such that \( x + y \) is divisible by 5 is **60**.

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